Write a printf or scanf statement for each of the following: Print unsigned integer 40000 left justified in a 15-digit field with 8 digits. Read a hexadecimal value into variable hex. Print 200 with and without a sign. Print 100 in hexadecimal form preceded by 0x. Read characters into array s until the letter p is encountered. Print 1.234 in a 9-digit field with preceding zeros. Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Use the assignment suppression character. Read a string of the form “characters” from the standard input. Store the string in character array s. Eliminate the quotation marks from the input stream. Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Do not use the assignment suppression character.

Question

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Answer 1

Textbook Question

Chapter 9, Problem 9.4E

Write a printf or scanf statement for each of the following:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.
Read a hexadecimal value into variable hex.
Print 200 with and without a sign.
Print 100 in hexadecimal form preceded by 0x.
Read characters into array s until the letter p is encountered.
Print 1.234 in a 9-digit field with preceding zeros.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Use the assignment suppression character.
Read a string of the form “characters” from the standard input. Store the string in character array s. Eliminate the quotation marks from the input stream.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Do not use the assignment suppression character.

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

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Q.1. Architecture performance [10 marks] Answer A certain microprocessor requires either 2, 4, or 6 machine cycles to perform various operations. ⚫ (40+g+f)% require 2 machine cycles, ⚫ (30-g) % require 4 machine cycles, and ⚫ (30-f)% require 6 machine cycles. (a) What is the average number of machine cycles per instruction for this microprocessor? Answer (b) What is the clock rate (machine cycles per second) required for this microprocessor to be a "1000 MIPS" processor? Answer (c) Suppose that 35% of the instructions require retrieving an operand from memory which needs an extra 8 machine cycles. What is the average number of machine cycles per instruction, including the instructions that fetch operands from memory?

Q.2. Architecture performance [25 marks] Consider two different implementations, M1 and M2, of the same instruction set. M1 has a clock rate of 2 GHz and M2 has a clock rate of 3.3 GHz. There are two classes of instructions with the following CPIs: Class A CPI for M1 CPI for M2 2.f 1.g B 5 3 C 6 4 Note that the dots in 2 fand 1.g indicate decimal points and not multiplication. a) What are the peak MIPS performances for both machines? b) Which implementation is faster, if half the instructions executed in a certain program are from class A, while the rest are divided equally among classes B and C. c) What speedup factor for the execution of class-A instructions would lead to 20% overall speedup? d) What is the maximum possible speedup that can be achieved by only improving the execution of class-A instructions? Explain why. e) What is the clock rate required for microprocessor M1 to be a "1000 MIPS" (not peak MIPS) processor?

Answer 2

Textbook Question

Chapter 9, Problem 9.4E

Write a printf or scanf statement for each of the following:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.
Read a hexadecimal value into variable hex.
Print 200 with and without a sign.
Print 100 in hexadecimal form preceded by 0x.
Read characters into array s until the letter p is encountered.
Print 1.234 in a 9-digit field with preceding zeros.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Use the assignment suppression character.
Read a string of the form “characters” from the standard input. Store the string in character array s. Eliminate the quotation marks from the input stream.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Do not use the assignment suppression character.

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 9, Problem 9.4E

Write a printf or scanf statement for each of the following:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.
Read a hexadecimal value into variable hex.
Print 200 with and without a sign.
Print 100 in hexadecimal form preceded by 0x.
Read characters into array s until the letter p is encountered.
Print 1.234 in a 9-digit field with preceding zeros.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Use the assignment suppression character.
Read a string of the form “characters” from the standard input. Store the string in character array s. Eliminate the quotation marks from the input stream.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Do not use the assignment suppression character.

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 9, Problem 9.4E

Write a printf or scanf statement for each of the following:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.
Read a hexadecimal value into variable hex.
Print 200 with and without a sign.
Print 100 in hexadecimal form preceded by 0x.
Read characters into array s until the letter p is encountered.
Print 1.234 in a 9-digit field with preceding zeros.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Use the assignment suppression character.
Read a string of the form “characters” from the standard input. Store the string in character array s. Eliminate the quotation marks from the input stream.
Read a time of the form hh:mm:ss, storing the parts of the time in the integer variables hour, minute and second. Skip the colons (:) in the input stream. Do not use the assignment suppression character.

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

Answer 8

a.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 13

Explanation of Solution

Given information:

Print unsigned integer 40000 left justified in a 15-digit field with 8 digits.

Explanation:

Following is the print statement to print an unsigned integer 40000 left justified in a 15-digit field with 8 digits:

printf ( “%-15.8u”,40000 ) ;

The printf displays the given number as 8 digits with left justification in a l5 digit field.

-15 placed to the immediate right of % sign, is used to left justify the number and occupy 15 spaces.
Digit 8 after the decimal point places zeros to the left of the number to make it an 8-digit number.
Conversion specifier u is used to print unsigned numbers.

Answer 14

b.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 19

Explanation of Solution

Given information:

Read a hexadecimal value into variable hex.

Explanation:

Following is the scanf statement to read a hexadecimal value into hex variable:

scanf ( “%x”, hex ) ;

The scanf statement inputs a hexadecimal number in the variable named hex using conversion specifier %x.

Answer 20

c.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

Answer 21

c.

Expert Solution

Answer 22

c.

Expert Solution

Answer 23

Expert Solution

Answer 24

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 25

Explanation of Solution

Given information:

Print 200 with and without a sign.

Explanation:

Following is the printf statement to print 200 with and without a sign:

printf ( “%+d\n %d\n”, 200, 200 ) ;

The printf statement prints the value 200, with and without a plus sign. If we place a + sign immediate to the right of %sign, then, a positive value is printed with a plus sign and a negative value is printed with a minus sign.

Answer 26

d.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

Answer 27

d.

Expert Solution

Answer 28

d.

Expert Solution

Answer 29

Expert Solution

Answer 30

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 31

Explanation of Solution

Given information:

Print 100 in hexadecimal form preceded by 0x.

Explanation:

Following is the printf statement to print 100 in hexadecimal form preceded by 0x:

printf ( “%#x\n”, 100 ) ;

The printf statement is used to print 100 preceded by Ox. This is accomplished by using a # flag placed immediate to the right of % sign in the field.

Answer 32

e.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

Answer 33

e.

Expert Solution

Answer 34

e.

Expert Solution

Answer 35

Expert Solution

Answer 36

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 37

Explanation of Solution

Given information:

Read characters into arrays s until the letter p is encountered.

Explanation:

Following is the scanf statement to read characters into arrays s until the letter p is encountered:

scanf ( “%[^p]”, s ) ;

The scanf statement is used to read the string until the letter p appears. This is achieved by inverted scan set, that is, by placing a Caret (^) before the character.

Answer 38

f.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

Answer 39

f.

Expert Solution

Answer 40

f.

Expert Solution

Answer 41

Expert Solution

Answer 42

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 43

Explanation of Solution

Given information:

Print 1.234 in a 9-digit field with preceding zeros.

Explanation:

Following is the printf statement to print 1.234 in a 9-digit field with preceding zeros:

printf ( “%09.3f\n”, 1.234 ) ;

The printf statement prints the given floating-point number in the field of 9 digits and preceded by zeros. Thus achieved by placing 09 immediate to the right of % sign.

Digit 3 after the decimal point is used to provide precision up to 3 values.

Answer 44

g.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for the given condition.

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

Answer 45

g.

Expert Solution

Answer 46

g.

Expert Solution

Answer 47

Expert Solution

Answer 48

Program Plan Intro

To write a printf or scanf statement for the given condition.

Answer 49

Explanation of Solution

Given information:

Write time of the hh: mm: ss type, storing the timepieces in the hour, minute and second integer variables. Skip the colons (:) through the input tube. Use the character assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss -:

scanf ( “%d*c%d*c%d”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated using the suppression character (*) in the field.

Answer 50

h.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

Answer 51

h.

Expert Solution

Answer 52

h.

Expert Solution

Answer 53

Expert Solution

Answer 54

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 55

Explanation of Solution

Given information:

Read a string from the standard input of the "characters" type. Place the string in character array s. Eliminate quotation marks from the input stream.

Explanation:

Following is the scanf statement to read a string of characters and store in array s -:

scanf ( “\”%[^\“]”, s ) ;

The scanf statement is used to input a string by the in quotation marks in a character array, s, and eliminate those quotation marks.

This is achieved by using an inverted scan set where a caret (^) before the \” sign is placed.

Answer 56

i.

Expert Solution

Program Plan Intro

To write a printf or scanf statement for given condition.

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.

Answer 57

i.

Expert Solution

Answer 58

i.

Expert Solution

Answer 59

Expert Solution

Answer 60

Program Plan Intro

To write a printf or scanf statement for given condition.

Answer 61

Explanation of Solution

Given information:

Read a time of type hh: mm: ss, storing time pieces in the hour, minute and second integer variables. Skip the input stream colons (:). Do not use the character Assignment Suppression.

Explanation:

Following is the scanf statement to read the time in form hh:mm:ss and skip the colons without using assignment suppression character -:

scanf ( “%d:%d:%d: ”, &hour, &minute, &second ) ;

The scanf statement is used to input the time in the form hh:mm:ss. The colons (:) are eliminated by placing colons (:) in the scanf statement as shown, if we do not have to put suppression character.