Variadic function (va_arg) doesn't work with float?

Question

I have a variadic function which takes a float parameter. Why doesn\'t it work?

va_arg(arg, float)

Answer 1

As @dasblinkenlight has mentioned, float is promoted to double. It works fine for me:

#include <stdio.h>          
#include <stdarg.h>

void foo(int n, ...)
{   
    va_list vl;
    va_start(vl, n);

    int c; 
    double val; 

    for(c = 0; c < n; c++) {
        val = va_arg(vl, double);
        printf("%f\n", val);
    }

    va_end(vl);
}


int main(void)
{
  foo(2, 3.3f, 4.4f);
  return 0;
}

Output:

3.300000
4.400000

Answer 2

Parameters of functions that correspond to ... are promoted before passing to your variadic function. char and short are promoted to int, float is promoted to double, etc.

6.5.2.2.7 The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

The reason for this is that early versions of C did not have function prototypes; parameter types were declared at the function site but were not known at the call site. But different types are represented differently, and the representation of the passed argument must match the called function's expectation. So that char and short values could be passed to functions with int parameters, or float values could be passed to functions with double parameters, the compiler "promoted" the smaller types to be of the larger type. This behavior is still seen when the type of the parameter is not known at the call site -- namely, for variadic functions or functions declared without a prototype (e.g., int foo();).