(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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Students have asked these similar questions

8.4 Self-Bias Configuration 20. Determine Zi. Zo. and A,, for the network of Fig. 8.73 if gf, = 3000 μS and gos = 50 μs. 21. Determine Z, Zo, and A, for the network of Fig. 8.73 if the 20-uF capacitor is removed and the parameters of the network are the same as in Problem 20. Compare results with those of Problem 20. +12 V 3.3 ΚΩ HE C₂ Vo Z Zo C₁ 10 ΜΩ Z₁ 1.1 ΚΩ Cs 20 µF FIG. 8.73 Problems 20, 21, 22, and 59.

21. Determine Zi, Zo, and A, for the network of Fig. 8.73 if the 20-μF capacitor is removed and the parameters of the network are the same as in Problem 20. Compare results with those of Problem 20. +12 V 3.3 ΚΩ +6 C₂ C₁ Z₁ 10 ΜΩ 1.1 ΚΩ Cs 20 μF FIG. 8.73 Zo

Ninth Edition Determine Zi, Zo and Av 20 V Zi + 1 ΜΩ 2 ΚΩ HH Z IDSS= 6MA Vp=-6V Yos = 40μS 20 and 47

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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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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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