(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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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

L et D= { s, d, b}, E= { c, k}, G= { a, k }, U={a, b, c, d, k, s} DUE=

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);

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)...

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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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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

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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