How to multiply float with integers in C?

Question

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

Delogu

It is a matter of integer and float - see my answer. – Rollet

@Carl is right. It may not be a float/int problem with the multiplication itself but it definitely is in the format string. – Microanalysis

The function printf is a varargs method, with signature int printf ( const char * format, ... ). This means that it takes arbitrary arguments after the first, and it cannot check them at compile time or cast them to a desired type taken from the format string. Your format string says %d, which looks for an int. The expression a*b is float, which gets converted to double. These types don't even have the same memory size, so the temporary double gets physically cut in half and treated as an int. This is not a Solomonic solution. – Fortyniner

possible duplicate of Arduino sprintf float not formatting – Pep

Answer 1

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

Rollet

Wow, %a must be in the running for the "most useless format specifier in a printf role" award :-) – Microanalysis

It's great! You can transfer (binary) floating point numbers 100% accurately via strings by using it, even to systems that don't use IEEE754. It preserves all of the bits of precision in a completely unambiguous way. Users probably don't care to see it, though.... – Rollet

Now that's a use case I hadn't considered. Good call. – Microanalysis

First time seing %a! Except for debug purpose, i dont really see the use for it. – Grasmere

Answer 2

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.

Bayle