(i). Given statement: //Initialization of char array char stringVar[10]="Hello"; Explanation: In the above given statement the string “Hello” is stored in an array as a series of characters. First, the character “H” is stored in “stringVar[0]”. The character “e” is stored in “stringVar[1]”. The character “l” is stored in “stringVar[2]”. The character “l” is stored in “stringVar[3]”. The character “o” is stored in “stringVar[4]”. Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”. (ii)...

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Answer 1

Expert Solution & Answer

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

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1. Look at the following partial code. function checker(x) { var got_an_A = true; if (x >= 90) { got_an_A = true; } else { got _an_A = false; if (got_an_A != true) { document.getElementById("tbx1").value = "Did not get the A"; } else { document.getElementByld("tbx1").value = "Got the A"; } a. What will be the value of the textbox with the id: thxl if we have the following function call: checker(95); b. What will be the value of the textbox with the id: thxl if we have the following function call: checker(85);

solve in c++ Create an application for a supermarket owner. The owner would like to storeup to 50 items. Each item is represented by an ID and a price.The program should ask the user if he/she wants to do one of the followingtasks:1. Enter an item ID and price2. Retrieve the price of a specific item3. Display all items within a specific range of prices4. Delete an item5. Display all items6. Delete all itemsNote: The items should be stored in memory sorted based on price. The six tasks should be implemented as functions.The program should run continuously until the user presses 'e' to exit

Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eax

Answer 2

Expert Solution & Answer

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Expert Solution & Answer

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Answer 3

Expert Solution & Answer

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Expert Solution & Answer

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See solution

Answer 4

Expert Solution & Answer

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Answer 8

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Answer 9

Chapter 8.1, Problem 1STE

Explanation of Solution

(i).

Given statement:

//Initialization of char array

char stringVar[10]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[10]”.

(ii)...

Answer 10

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Answer 11

Explanation of Solution

(iii).

Given statement:

// Initialization of char array

char stringVar[10]={'H','e','l','l','o'};

Explanation:

In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”...

Answer 12

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Answer 13

Explanation of Solution

(iv).

Given statement:

// Initialization of char array

char stringVar[6]="Hello";

Explanation:

In the above given statement the string “Hello” is stored in an array as a series of characters.

First, the character “H” is stored in “stringVar[0]”.

The character “e” is stored in “stringVar[1]”.

The character “l” is stored in “stringVar[2]”.

The character “l” is stored in “stringVar[3]”.

The character “o” is stored in “stringVar[4]”.

Finally, the null character “\0” is stored in “stringVar[5]”. So, it print the result “Hello” because the size of array is “stringVar[6]”.

(v)...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 8.1 - Prob. 1STE Ch. 8.1 - What C string will be stored in singingString...Ch. 8.1 - What (if anything) is wrong with the following...Ch. 8.1 - Suppose the function strlen (which returns the...Ch. 8.1 - Prob. 5STE Ch. 8.1 - How many characters are in each of the following...Ch. 8.1 - Prob. 7STE Ch. 8.1 - Given the following declaration and initialization...Ch. 8.1 - Given the declaration of a C-string variable,...Ch. 8.1 - Write code using a library function to copy the...

Solution Summary: The author explains that the characters "Hello" are stored in an array as a series of characters.

Explanation of Solution

Explanation of Solution

Explanation of Solution

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Chapter 8 Solutions

Solution Summary: The author explains that the characters "Hello" are stored in an array as a series of characters.

Explanation of Solution

Explanation of Solution

(iii). Given statement: // Initialization of char array char stringVar[10]={'H','e','l','l','o'}; Explanation: In the above given statement the characters “{'H','e','l','l','o'}”are stored in an array as a series of characters. First, the character “H” is stored in “stringVar[0]”...

Explanation of Solution

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Chapter 8 Solutions