Вопрос

When I execute this code it returns me 1610612736

void main(){
float a=3.3f;
int b=2;
printf("%d",a*b);
}

Why and how to fix this ?

edit : It's not even a matter of integer and float, if i replace int b=2: by float b=2.0f it return the same silly result

Это было полезно?

Решение

The result of the multiplication of a float and an int is a float. Besides that, it will get promoted to double when passing to printf. You need a %a, %e, %f or %g format. The %d format is used to print int types.

Editorial note: The return value of main should be int. Here's a fixed program:

#include <stdio.h>

int main(void)
{
    float a = 3.3f;
    int b = 2;
    printf("%a\n", a * b);
    printf("%e\n", a * b);
    printf("%f\n", a * b);
    printf("%g\n", a * b);
    return 0;
}

and its output:

$ ./example 
0x1.a66666p+2
6.600000e+00
6.600000
6.6

Другие советы

Alternately, you could also do

printf("%d\n", (int)(a*b));

and this would print the result you're (kind of) expecting.

You should always explicitly typecast the variables to match the format string, otherwise you could see some weird values printed.

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