Pointers · Exercise 4 of 4

Pointers Pointer Programs

8 questions
Page 1 of 1
Answers explained

Read the concepts Take a timed test

Your progressThis exercise · saved in this browser

Attempted: 0/8
Correct: 0
Accuracy: –

Continue practice

OverviewExercise at a glance

Category: C Programming
Topic: Pointers
Exercise: Pointer Programs
Questions: 8

Exercise list4 exercises in Pointers

Study materialRevise before you answer

Quick tipsPractise smarter

Solve it yourself before opening the answer
Use the workspace for rough steps
Bookmark questions to revise later
Aim for under a minute per question

0Attempted

0Correct

0Wrong

Directions

Write each program yourself first, then open the solution. Every solution was compiled and run; the output shown is the real output. Tick "Mark as solved" to track your progress.

Question 1 Easy

Write a function swap() that swaps two integers using pointers, and call it from main().

Solution

main.c

#include <stdio.h>

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int main(void) {
    int a = 10, b = 20;
    printf("Before: a = %d, b = %d\n", a, b);
    swap(&a, &b);
    printf("After:  a = %d, b = %d\n", a, b);
    return 0;
}

Output Verified by compiling and running

Before: a = 10, b = 20
After:  a = 20, b = 10

The function receives the addresses of a and b, so writing through *a and *b changes the caller's variables. A swap that takes plain ints would only swap its own copies.

Question page & discussion

Question 2 Easy

Find the sum and the average of the elements of an integer array by moving a pointer through it (no a[i]).

Solution

main.c

#include <stdio.h>

int main(void) {
    int marks[] = {78, 92, 65, 88, 70};
    int n = sizeof marks / sizeof marks[0];
    int sum = 0;

    for (int *p = marks; p < marks + n; p++)
        sum += *p;

    printf("Sum = %d\n", sum);
    printf("Average = %.2f\n", (double) sum / n);
    return 0;
}

Output Verified by compiling and running

Sum = 393
Average = 78.60

marks + n is the address just past the last element, a safe end marker that is never dereferenced. Casting to double before dividing keeps the decimals.

Question page & discussion

Question 3 Medium

Reverse an integer array in place using two pointers, one from each end.

Solution

main.c

#include <stdio.h>

void reverse(int *a, int n) {
    int *left = a, *right = a + n - 1;
    while (left < right) {
        int t = *left;
        *left++ = *right;
        *right-- = t;
    }
}

int main(void) {
    int a[] = {1, 2, 3, 4, 5, 6};
    int n = sizeof a / sizeof a[0];
    reverse(a, n);
    for (int i = 0; i < n; i++)
        printf("%d ", a[i]);
    printf("\n");
    return 0;
}

Output Verified by compiling and running

6 5 4 3 2 1

Swap the elements at the two pointers, then move them towards each other until they meet: n/2 swaps, no second array. *left++ = *right stores, then moves left forward.

Question page & discussion

Question 4 Easy

Write my_strlen() that returns the length of a string using a pointer, without calling strlen().

Solution

main.c

#include <stdio.h>

int my_strlen(const char *s) {
    const char *p = s;
    while (*p != '\0')
        p++;
    return (int) (p - s);
}

int main(void) {
    const char *words[] = {"C", "pointer", "LearnFrenzy", ""};
    for (int i = 0; i < 4; i++)
        printf("length of \"%s\" = %d\n", words[i], my_strlen(words[i]));
    return 0;
}

Output Verified by compiling and running

length of "C" = 1
length of "pointer" = 7
length of "LearnFrenzy" = 11
length of "" = 0

Move p to the terminating '\0'; the distance p - s is the number of characters before it. The empty string gives 0 because the loop never runs.

Question page & discussion

Question 5 Medium

Sort an integer array in ascending order with bubble sort, accessing the elements only through pointers.

Solution

main.c

#include <stdio.h>

void sort(int *a, int n) {
    for (int i = 0; i < n - 1; i++)
        for (int *p = a; p < a + n - 1 - i; p++)
            if (*p > *(p + 1)) {
                int t = *p;
                *p = *(p + 1);
                *(p + 1) = t;
            }
}

int main(void) {
    int a[] = {42, 7, 19, 3, 25};
    int n = sizeof a / sizeof a[0];
    sort(a, n);
    for (int *p = a; p < a + n; p++)
        printf("%d ", *p);
    printf("\n");
    return 0;
}

Output Verified by compiling and running

3 7 19 25 42

After pass i the largest i + 1 values are already at the end, so the inner pointer stops at a + n - 1 - i. Neighbours *p and *(p + 1) are swapped when they are out of order.

Question page & discussion

Question 6 Easy

Count the vowels and consonants in a sentence by walking through it with a char pointer.

Solution

main.c

#include <stdio.h>
#include <ctype.h>

int main(void) {
    const char *s = "Pointers make C powerful";
    int vowels = 0, consonants = 0;

    for (const char *p = s; *p; p++) {
        char c = (char) tolower((unsigned char) *p);
        if (c >= 'a' && c <= 'z') {
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
                vowels++;
            else
                consonants++;
        }
    }
    printf("Vowels: %d\nConsonants: %d\n", vowels, consonants);
    return 0;
}

Output Verified by compiling and running

Vowels: 8
Consonants: 13

The loop runs until *p is the terminating zero. tolower() handles capital letters, and only letters are counted, so spaces are skipped.

Question page & discussion

Question 7 Medium

Read n, allocate an array of n integers with malloc(), read the numbers, and print their sum, maximum and average. Free the memory at the end.

Input: 5 then 12 45 7 30 6

Solution

main.c

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

int main(void) {
    int n;
    if (scanf("%d", &n) != 1 || n <= 0)
        return 1;

    int *a = malloc(n * sizeof *a);     /* room for exactly n ints */
    if (a == NULL)
        return 1;

    for (int i = 0; i < n; i++)
        if (scanf("%d", a + i) != 1)    /* a + i is the address of a[i] */
            return 1;

    int sum = 0, max = a[0];
    for (int *p = a; p < a + n; p++) {
        sum += *p;
        if (*p > max)
            max = *p;
    }
    printf("Sum = %d, Max = %d, Average = %.1f\n", sum, max, (double) sum / n);
    free(a);
    return 0;
}

Output Verified by compiling and running

Sum = 100, Max = 45, Average = 20.0

malloc(n * sizeof *a) asks for exactly n ints, and writing sizeof *a keeps the size right if the type changes. a + i is already an address, so scanf() needs no &. Every malloc() is paired with a free().

Question page & discussion

Question 8 Medium

Write my_strcpy() that copies one string into another using pointers, without calling strcpy().

Solution

main.c

#include <stdio.h>

void my_strcpy(char *dest, const char *src) {
    while ((*dest++ = *src++) != '\0')
        ;
}

int main(void) {
    char copy[20];
    my_strcpy(copy, "LearnFrenzy");
    printf("Copied: %s\n", copy);
    return 0;
}

Output Verified by compiling and running

Copied: LearnFrenzy

Each loop step copies one character and moves both pointers. The loop stops after copying the terminating '\0', so the copy is a proper string. The destination must be large enough – here 20 bytes for 12 characters.

Question page & discussion

All exercises doneBack to PointersRevise the concepts or pick another topic

Ready to test your Programming?

Timed online tests with instant scores and explanations.