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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#include using namespace std; int find(int arr[], int value, int left, int right) { int midpt = (left+right)/2; if (left > right) return -1; if ( arr[midpt] return midpt; else if (arr[midpt] < value) == value) return else return find(arr,value, left,midpt-1); } void main(void) { int arr[] ={4,5,6,9,11}; cout<

#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000

String Pair // Problem Description // One person hands over the list of digits to Mr. String, But Mr. String understands only strings. Within strings also he understands only vowels. Mr. String needs your help to find the total number of pairs which add up to a certain digit D. // The rules to calculate digit D are as follow // Take all digits and convert them into their textual representation // Next, sum up the number of vowels i.e. {a, e, i, o, u} from all textual representation // This sum is digit D // Now, once digit D is known find out all unordered pairs of numbers in input whose sum is equal to D. Refer example section for better understanding. // Constraints // 1 <= N <= 100 // 1 <= value of each element in second line of input <= 100 // Number 100, if and when it appears in input should be converted to textual representation as hundred and not as one hundred. Hence number…

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

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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

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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...