Infinity in MSVC++
I'm using MSVC++, and I want to use the special value INFINITY in my code.
What's the byte patt开发者_JS百科ern or constant to use in MSVC++ for infinity?
Why does 1.0f/0.0f appear to have the value 0?
#include <stdio.h>
#include <limits.h>
int main()
{
float zero = 0.0f ;
float inf = 1.0f/zero ;
printf( "%f\n", inf ) ; // 1.#INF00
printf( "%x\n", inf ) ; // why is this 0?
printf( "%f\n", zero ) ; // 0.000000
printf( "%x\n", zero ) ; // 0
}
Use numeric_limits
:
#include <limits>
float maxFloat = std::numeric_limits<float>::infinity();
printf("%x\n", inf)
expects an integer (32 bit on MSVC), but receives a double. Hilarity will ensue. Err, I mean: undefined behavior.
(And yes, it receives a double since for a variable argument list, floats are promoted to double).
Edit anyways, you should use numeric_limits
, as the other reply says, too.
In the variable arguments list to printf, floats get promoted to doubles. The little-endian byte representation of infinity as a double is 00 00 00 00 00 00 F0 7F.
As peterchen mentioned, "%x" expects an int, not a double. So printf looks at only the first sizeof(int) bytes of the argument. No version of MSVC++ defines int to be larger than 4 bytes, so you get all zeros.
Take a look at numeric_limits::infinity
.
That's what happens when you lie to printf(), it gets it wrong. When you use the %x format specifier, it expects an integer to be passed on the stack, not a float passed on the FPU stack. Fix:
printf( "%x\n", *(__int32*)&inf ) ;
You can get infinity out of the <limits>
C++ header file:
float inf = std::numeric_limits<float>::infinity().
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