CODEWITHSUNDEEP
c
MyGanton
MyGanton

Reputation: 37

Define a function to check if its a perfect square in C

I tried to write a code to check if a number is a perfect square, but I'm not able to call the function I defined. Where is my mistake?

#include <stdio.h>
#include <math.h>

int isPerfectSquare(int number) {
    int i;
    for (i = 0; i <= number; i++) {
        if (number == (i * i)) {
            printf("Success");
            break;
        } else {
            continue;
        }
    }
    printf("Fail");
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", n);
    isPerfectSquare(n);
    return 0;
}

I don't get any answer ("Success" or "Fail").

Upvotes: 1

Views: 1294

Answers (1)

chqrlie
chqrlie

Reputation: 144540

You must pass the address of n instead of its value in scanf("%d", n);:

scanf("%d", &n);

Note however that your function will print both Success and Fail for perfect squares because you should return from the function instead of just breaking from the loop upon success.

Here is a modified version:

#include <stdio.h>

void isPerfectSquare(int number) {
    int i;
    for (i = 0; i <= number; i++) {
        if (number == (i * i)) {
            printf("Success\n");
            return;
        }
    }
    printf("Fail\n");
}

int main() {
    int n;
    printf("Enter a number: ");
    if (scanf("%d", &n) == 1) {
        isPerfectSquare(n);
    }
    return 0;
}

Note also that your method is quite slow and may have undefined behavior (and produce false positives) if i becomes so large that i * i exceeds the range of type int. You should instead use a faster method to figure an approximation of the square root of n and check if the result is exact.

It is also better for functions such as isPerfectSquare() to return a boolean value instead of printing some message, and let the caller print the message. Here is a modified version using the Babylonian method, also known as Heron's method.

#include <stdio.h>

int isPerfectSquare(int number) {
    int s1 = 2;
    if (number < 0)
        return 0;
    // use the Babylonian method with 10 iterations
    for (int i = 0; i < 10; i++) {
        s2 = (s1 + number / s1) / 2;
        if (s1 == s2)
            break;
        s1 = s2;
    }
    return s1 * s1 == number;
}

int main() {
    int n;
    printf("Enter a number: ");
    if (scanf("%d", &n) == 1) {
        if (isPerfectSquare(n)) {
            printf("Success\n");
        } else {
            printf("Fail\n");
        }
    }
    return 0;
}

Upvotes: 2

Related Questions