SEBA Class 10 Computer Science Textbook Solution Chapter 8

 

Explore the solutions chapter-wise:

SL. No.	Contents
Chapter 1	Introduction to Computer Network
Chapter 2	HTML5 and CSS3
Chapter 3	Database Part - II MySQL
Chapter 4	Introduction to Loops
Chapter 5	Nested Loops in C
Chapter 6	Arrays in C
Chapter 7	Functions in C
Chapter 8	Pointers in C
Chapter 9	Structure in C
Chapter 10	An Introduction to Object Oriented Programming
Chapter 11	Case Studies

Chapter 8 – Pointers in C

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Short Answer Type Questions

1. How is a pointer variable different from a normal variable?
Ans: A normal variable stores a value, such as an integer or a character.
A pointer variable stores the address of another variable. Just like normal variables, pointers must be declared before use.

Example:

a) int x = 10; → stores value 10  
b) int *p = &x; → stores the address of x

2. Why is dynamic memory allocation an efficient memory management technique?
Ans: Dynamic memory allocation allows memory to be allocated during program execution and freed when not needed.
This helps in:
a) Better use of memory
b) Avoiding memory wastage
c) Reusing memory when required

Hence, dynamic memory allocation is an efficient way to manage memory.

3. How many bytes are needed to store an int pointer variable? Is it the same for a char pointer variable?
Ans: An int pointer variable requires 4 bytes of memory.
Yes, a char pointer variable also requires 4 bytes.
Pointer size depends on the system architecture, not on the data type.

4. Write the output of the following code segments.

(a)

int *ptr, x = 9;
ptr = &x;
printf("%d", (*ptr)++);

Output:
9

(b)

int *ptr, x = 9;
ptr = &x;
printf("%d", (*ptr)++);
printf("%d", *ptr);

Output:
9
10

(c)

int *ptr, x = 9;
ptr = &x;
int y = ++(*ptr);
printf("%d", y);

Output:
10

(d)

char *ptr, x = 'A';
ptr = &x;
char y = *ptr;
printf("%c", y);

Output:
A

(e)

char *ptr, x = 'A';
ptr = &x;
char y = (*ptr)++;
printf("%c", y);

Output:
A

(f)

char *ptr, x = 'A';
ptr = &x;
char y = ++(*ptr);
printf("%c", y);

Output:
B

(g)

char *ptr, x = 'A';
ptr = &x;
char *y;
y = ptr;
printf("%c", ++(*y));

Output:
B

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Long Answer / Programming Questions

5. Program: Dynamic memory allocation for array

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int *a, i;
    a = (int *)malloc(10 * sizeof(int));

    for (i = 0; i < 10; i++)
        a[i] = i + 1;

    for (i = 0; i < 5; i++)
        printf("%d ", a[i]);

    free(a);
    return 0;
}

6. Program: Maximum runs using dynamic memory

#include <stdio.h>
#include <stdlib.h>

int max(int *a)
{
    int m = a[0];
    for (int i = 1; i < 10; i++)
        if (a[i] > m)
            m = a[i];
    return m;
}

int main()
{
    int *a;
    a = (int *)malloc(10 * sizeof(int));

    for (int i = 0; i < 10; i++)
        scanf("%d", &a[i]);

    printf("Maximum run = %d", max(a));

    free(a);
    return 0;
}

7. Program: Dynamic memory for name

#include <stdio.h>
#include <stdlib.h>

char *create(int n)
{
    return (char *)malloc((n + 1) * sizeof(char));
}

void show(char *s)
{
    printf("%s", s);
}

int main()
{
    int n;
    char *name;

    scanf("%d", &n);
    name = create(n);
    scanf("%s", name);

    show(name);
    free(name);
    return 0;
}

8. Program: Even numbers using dynamic memory

#include <stdio.h>
#include <stdlib.h>

int countEven(int *a, int n)
{
    int c = 0;
    for (int i = 0; i < n; i++)
        if (a[i] % 2 == 0)
            c++;
    return c;
}

int *copyEven(int *a, int n, int c)
{
    int *b = (int *)malloc(c * sizeof(int));
    int j = 0;
    for (int i = 0; i < n; i++)
        if (a[i] % 2 == 0)
            b[j++] = a[i];
    return b;
}

int main()
{
    int a[] = {1,2,3,4,5,6,7,8,9,10};
    int c = countEven(a, 10);
    int *b = copyEven(a, 10, c);

    for (int i = 0; i < c; i++)
        printf("%d ", b[i]);

    free(b);
    return 0;
}

9. Program: Numbers divisible by 3

#include <stdio.h>
#include <stdlib.h>

int count(int *a, int n)
{
    int c = 0;
    for (int i = 0; i < n; i++)
        if (a[i] != 0 && a[i] % 3 == 0)
            c++;
    return c;
}

int *copy(int *a, int n, int c)
{
    int *b = (int *)malloc(c * sizeof(int));
    int j = 0;
    for (int i = 0; i < n; i++)
        if (a[i] != 0 && a[i] % 3 == 0)
            b[j++] = a[i];
    return b;
}

int sum(int *a, int n)
{
    int s = 0;
    for (int i = 0; i < n; i++)
        s += a[i];
    return s;
}

int main()
{
    int a[] = {1,0,9,12,6,15,0,18,3,0};
    int c = count(a, 10);
    int *b = copy(a, 10, c);

    printf("Sum = %d", sum(b, c));

    free(b);
    return 0;
}