Value assigned to s variable when uses %d instead %f [closed] - c

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I assume output to be 8 but the line s=a+b assign value 5 to s variable not sum of a+b. I know that I am using %d instead of %f.
#include <stdio.h>
void summ(int, int);
int main(){
int a = 3, b = 5;
summ(a, b);
return 0;
}
void summ(int a, int b){
float s;
s = a + b;
printf("%d", s);
}

Answer 1

You can't use any old format specifier in a printf(). It always has to match the type of the variable you are passing in. If you want to print a value as another type, you have to convert it first, then print the converted value. So either this has to be:
void summ(int a, int b)
{
float s;
s = a + b;
printf( "%f\n", s );
}
or
void summ(int a, int b)
{
int s;
s = a + b;
printf( "%d\n", s );
}
In the case of your sum function, you won't see a noticeable difference (except that %d does not print a ".0000" after the number, but you could have suppressed that by adding a length of 0 to the fraction of your %f by writing it as %.0f).
OTOH, if you have a division:
void divv(int a, int b)
{
float s;
s = a / b;
printf( "%f\n", s );
}
the CPU will perform a (often faster) integer division. That means that 3 / 5 (which would be 0.6) just gives you 0 because it didn't bother calculating the fraction. The compiler doesn't look at the type of variable an expression is assigned to, so it doesn't see that it could actually store the fraction. It just looks at the operands of the operator, which are both int, calculates an int, and expands it into a float. So if you wanted a precise result, you'd have to do:
void divv(int a, int b)
{
float s;
s = (float)a / (float)b;
printf( "%f\n", s );
}
because as soon as one of the arguments is a float, it will perform a floating point division and also calculate the fraction.

Answer 2

here is a possible version of the code that cleanly compiles and performs the desired functionality.
#include <stdio.h>
void summ(int, int);
int main( void )
{
int a = 3;
int b = 5;
summ(a, b);
return 0;
} // end function: main
void summ(int a, int b)
{
float s;
s = (float)(a + b);
//printf("%d", s);
printf( "%f\n", s );
} // end function: summ
The output from the above code is:
8.000000