INFO: Type float Variable Arguments Are Promoted to Type double

• Microsoft C for MS-DOS, versions 5.1, 5.1a, 6.0, 6.0a, 6.0ax
• Microsoft C for OS/2, versions 5.1, 6.0, 6.0a
• Microsoft C/C++ for MS-DOS, version 7.0
• Microsoft Visual C++ for Windows, 16-bit edition, versions 1.0, 1.5
• Microsoft Visual C++, 32-bit Editions, versions 1.0, 2.0, 4.0, 5.0, 6.0

SUMMARY

When passing arguments of type float to a function that accepts a variable number of arguments, those floats not explicitly declared in the prototype will be promoted to type double.

In this case, the macros va_start and va_arg will use sizeof(float) to determine the argument size, when the actual size is sizeof(double). This will give unexpected results because the values on the stack will be interpreted incorrectly.

MORE INFORMATION

The sample program below returns the sum of a variable list of floating- point values. The sum returned is not correct because the variables of type float (other than the float that was declared) are promoted to double, but are treated as floats by va_start and va_arg.

Although this situation causes unintended results, this is expected behavior that conforms to the ANSI standard. Section 3.3.2.2 of the ANSI standard states:

If the expression that denotes the called function has a type that does not include a prototype, the integral promotions are performed on each argument and arguments that have type float are promoted to double. These are called the default argument promotions. . . . If the expression that denotes the called function has a type that includes a prototype, the arguments are implicitly converted to the types of the corresponding parameters. 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.

Sample Code

/* Compile options needed: none
*/ 

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

float sum( int num, float first, ... );

void main()
{
   float total;

   total = sum( 3, 11.11F, 22.22F, 33.33F );
   printf( "11.11 + 22.22 + 33.33 = %f\n\n", total );
}

float sum( int num, float first, ... )
{
   int count;
   float sum, temp = first;
   va_list marker;

   sum = 0.0F;
   va_start( marker, first );
   for( count=1; count<num; count++ )
   {
       sum += temp;
       temp = va_arg( marker, float );
   }
   sum += temp;

   va_end( marker );
   return( sum );
} 

Possible Program Output

   11.11 + 22.22 + 33.33 = -321864398408282700000000000.000000 

Additional query words:

Keywords : kbcode kbLangC kbVC100 kbVC150 kbVC200 kbVC400 kbVC500 kbVC600
Version : MS-DOS:5.1,6.0,6.00a,7.0; OS/2:5.1,6.0,6.00a; WINDOWS:1.0,1.5; WINDOWS NT:1.0,2.0,4.0,5.0
Platform : MS-DOS NT OS/2 WINDOWS
Issue type : kbinfo