floating-point constant comparison - (0.0 ? 1 : 0)

Question

In the example below, if uncomment float f = 0.0; ,
and replacing the return(0.0 ? 1 : 0); with return(f ? 1 : 0); .
The output is NIL .

Here is my code:

/* file main.c 
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 15.00.30729.01 for 80x86
cl -W4 -MTd -O2 -TC main.c -Fetest */   
#include <stdio.h>    
int my_func(void)
{
   /* float f = 0.0; */
   return(0.0 ? 1 : 0);
}
int main(void)
{  
    printf("%s\n", ( my_func() ? "ONE" : "NIL") );
    return 0;
}

On a 32-bit Windows machine, using Visual Studio this code outputs :

ONE

• Why my_func() returns value true (1) ?
• How does the C compiler interpret this expression (0.0 ? 1 : 0) ?

Answer 1

This looks like a bug in the Microsoft compiler which you should submit to Connect . I was able to duplicate it in Visual Studio Express 2010, but not in gcc: http://ideone.com/8qPRJd .

Any expression that evaluates to an integer value of 0 should be equivalent to false . This is exactly how it's working with the float variable, and it's the same when I tried it with a double as well.

Answer 2

return（0.0？1：0）编译为固定返回1.在另一种情况下，实际评估浮点变量，0.0不等于0。