Why does using int instead of float as a parameter give me a wrong result

Question

Whenever I define the parameter i in the function exam as an integer it always prints Sum = 0.00.

I would like to know why it is doing that...

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

float exam(float i) {
    if (i == 0) {
        return 0;
    } else {
        return (i / (i + 1)) + exam(i - 1);
    }
}

int main() {
    int i;
    printf("Enter an integer: ");
    scanf("%d", &i);
    printf("Sum = %.3f", exam(i));
    return 0;
}

don't forget about negative inputs – mukunda May 29 '22 at 22:10 — mukunda, May 29 '22 at 22:10

chqrlie · Answer 1 · 2022-05-29T22:30:19.007

If you define i as an int, the first part of the expression (i / (i + 1)) + exam(i - 1) is evaluated using integer arithmetics, hence i / (i + 1) has the value 0 and every recursive call adds 0 until i reaches 0, so the final result is 0.

Note also that the code would have undefined behavior for negative inputs as one of the calls would pass -1 and i / (i + 1) would cause a division by zero which has undefined behavior and usually causes program termination.

Another invalid input is INT_MAX for which i + 1 causes a arithmetic overflow.

Defining i as a float removes these shortcomings and computes the expected result. Using double instead of float for both the argument and return type would increase the precision of the computation.

@Rix: you can accept this answer eventhough the question has been closed. — chqrlie, May 29 '22 at 22:32

Why does using int instead of float as a parameter give me a wrong result

1 Answers1