Why is FLT_MIN equal to zero?

问题:

limits.h specifies limits for non-floating point math types, e.g. INT_MIN and INT_MAX. These values are the most negative and most positive values that you can represent using an int.

In float.h, there are definitions for FLT_MIN and FLT_MAX. If you do the following:

NSLog(@"%f %f", FLT_MIN, FLT_MAX);

You get the following output:

FLT_MIN = 0.000000, FLT_MAX = 340282346638528859811704183484516925440.000000

FLT_MAX is equal to a really large number, as you would expect, but why does FLT_MIN equal zero instead of a really large negative number?

回答1:

It's not actually zero, but it might look like zero if you inspect it using printf or NSLog by using %f.
According to float.h (at least in Mac OS X 10.6.2), FLT_MIN is described as:

/* Minimum normalized positive floating-point number, b**(emin - 1).  */

Note the positive in that sentence: FLT_MIN refers to the minimum (normalized) number greater than zero. (There are much smaller non-normalized numbers).

If you want the minimum floating point number (including negative numbers), use -FLT_MAX.

回答2:

The '%f' format prints 6 decimal places in fixed format. Since FLT_MIN is a lot smaller, it looks like zero in fixed point. If you use '%e' or '%g' format, you'd get a better formatted answer. Similarly with the FLT_MAX.

#include <float.h>
#include <stdio.h>
int main(void)
{
    printf("MIN = %f, MAX = %f\n", FLT_MIN, FLT_MAX);
    printf("MIN = %e, MAX = %e\n", FLT_MIN, FLT_MAX);
    return(0);
}


MIN = 0.000000, MAX = 340282346638528859811704183484516925440.000000
MIN = 1.175494e-38, MAX = 3.402823e+38