Pointers: A variable whose value is another variable’s address is termed as pointer. A pointer is to be declared before being used. The declaration syntax for a variable that is a pointer is shown below: type *lvarname; Here, “type ” is a pointer's base type and “lvarname ” is pointer variable’s name. The asterisk sign is been used for pointer declaration. The asterisk sign is being used for designation of a variable as a pointer.

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Answer 1

Question

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

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Answer 2

Question

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Expert Solution & Answer

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Answer 3

Question

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Answer 6

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Answer 7

Chapter 1, Problem 2E

Program Plan Intro

Pointers:

A variable whose value is another variable’s address is termed as pointer.

A pointer is to be declared before being used.

The declaration syntax for a variable that is a pointer is shown below:

type *lvarname;

Here, “type” is a pointer's base type and “lvarname” is pointer variable’s name.

The asterisk sign is been used for pointer declaration.

The asterisk sign is being used for designation of a variable as a pointer.

Answer 8

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Answer 9

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Answer 10

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Answer 11

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 1 - Prob. 1E Ch. 1 - Prob. 2E Ch. 1 - Prob. 3E Ch. 1 - Prob. 4E Ch. 1 - Prob. 5E Ch. 1 - Prob. 6E Ch. 1 - Prob. 7E Ch. 1 - Prob. 8E Ch. 1 - Prob. 9E Ch. 1 - Prob. 10E

Solution Summary: The author explains the purpose of a pointer, which is to save memory space and speed up execution time.

Concept explainers

Explanation of Solution

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

Explanation of Solution

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Want to see the full answer?

Chapter 1 Solutions

Solution Summary: The author explains the purpose of a pointer, which is to save memory space and speed up execution time.

Concept explainers

Explanation of Solution

b. Identification of errors: Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below: ...

Explanation of Solution

c. Identification of errors: While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...

Want to see the full answer?

Chapter 1 Solutions

b.

Identification of errors:

Before copying “s2” to “s1”, memory has to be allocated for accommodating contents of “s1”, as shown below:

...

c.

Identification of errors:

While memory allocation for contents of “s2” is performed, one more character has to be allocated for the end-of-string character. In the given syntax, one more character allocation is needed...