(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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this module is java 731 , follow all instructions and make sure the outputs are like what they expect and make sure the code is 100% correct . include all comments , layout and structure to be perfect too, thanks. Question 1: E-Hailing Bicycle Management System Case Study:An e-hailing company that rents out bicycles needs a system to manage its bicycles, users, and borrowing process. Each user can borrow up to 2 bicycles at a time, specifically for families with children 18 years or below. The system must track the bicycles (name, make, type, and availability) and users (name, ID, and borrowed bicycles). The company also wants to ensure that the system uses a multidimensional array to store information about the bicycles. Requirements: Add and View Bicycles: Borrow Bicycles: Return Bicycles Display Borrowed Bicycles and Search for a bicycle Create a menu-driven program to implement the above. Sample Output: Add Bicycle View All Bicycles Borrow Bicycle Return…

Add a method called transfer () to the BankAccount class, that takes in an amount and destinationAccount as input and transfer the funds from current account to destination account. This method should return the balance of the current account after the transfer, also should check for the sufficient balance in current account before proceeding the transfer and if there is insufficient balance return an error message "Insufficient balance". Modify the BankAccount Test class, so that it calls the transfer () method and prints the balance after transfer. Make sure that the transfer() method updates the balance of both the current account and the destination account.

• Create a public method called deposit() that takes in an amount of type double as input and adds the amount to the current balance. This method should also return the updated balance. • Create another public method called withdraw() that takes in an amount of type double as input, checks if the withdrawal amount is less than the current balance, and if so, subtracts the amount from the balance. If the withdrawal amount is greater than the current balance, the method should return an error message "Insufficient balance". • Create a public method getAccountInfo() that returns the account information in the format "Account Number: xxxxx, Account Holder: John Doe, Account Type: SAVINGS/CHECKING, Balance: $xxxX.XX". • Finally, create a constructor method that takes in the account number, account holder name, initial balance and account type as input and initializes the corresponding instance variables. In the BankAccount class, make sure to use the private access modifier for the instance…

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)...

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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)...

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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

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Answer 15

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