printf - Printing int as float in C

Question

I'm using Visual Studio TC compiler for Little Endian. The following is the piece of code:

void main()
{    
    float c = 1.0;
    int a = 0x3F800000;
    int *ptr = (int *)&c;
    printf("
0x%X
", *ptr);
    printf("
a = %f", a);
    printf("
c = %f", c);
    return;    
}

The output is:

0x3F800000
a = 0.000000
c = 1.000000

Float value 1.0 is 0x3F800000 and stored as 00 00 80 3F in memory for Little Endian. The same value is assigned to int a. How printf prints 0.000000 for int a while 1.000000 for float c? I've seen it prints all integer values as 0.000000, when printed using %f in printf.

Also, since printf is variable argument function, how does it knows whether the passed value in the register is int or float?

Answer 1

My psychic powers tell me Adam Liss's comment is the right answer: float arguments are promoted to double, so the printf() function expects that to happen: It expects a 64-bit value on the stack, but gets 32 bits plus garbage data that happen to be zero.

If you increase the display precision, the display should be something like a = 0.00000000001.

This also means this should work:

void main()
{    
    double c = 1.0;
    long long a = 0x3FF0000000000000;
    long long *ptr = (long long *)&c;
    printf("
0x%llX
", *ptr);
    printf("
a = %f", a);
    printf("
c = %f", c);
    return;    
}