Floating Point subnormal numbers under C99 with GCC 4.7‏

[Argentinator Rincón Matemático]

Hi, dear friends.

I am testing floating-points macros in C language, under the standard C99.
My compiler is GCC 4.6.1. (with 4.7.1, I have the same result).

I have two computers:
My system (1) is Windows XP SP2 32bit, in an "Intel (R) Celeron (R) 420" @ 1.60 GHz.
My system (2) is Windows 7 Ultimate SP1 64bit, in an "AMD Turion II X2 dual-core mobile M520 ( 2,3 ghz 1MB L2 Cache )"
(The result was the same in both systems.)

I am interested in testing subnormal numbers for the types float, double and long double.
I've tried the following line:

printf(" Float: %x\n Double: %x\n Long Double: %x\n",fpclassify(FLT_MIN / 4.F), fpclassify(DBL_MIN / 4.), fpclassify(LDBL_MIN / 4.L ));

I've compiled with the options -std=c99 and -pedantic (also without -pedantic).
Compilation goes well, however the program shows me this:

 Float: 400
 Double: 400
 Long Double: 4400

(0x400 == FP_NORMAL, 0x4400 == FP_SUBNORMAL)

I think that the right result must be 0x4400 in all cases.

When I tested the constant sizes, I have obtained they are of the right type.
For example, I have obtained:

sizeof(float) == 4
sizeof(double) == 8
sizeof(long double) == 12

Also:

sizeof(FLT_MIN / 4.F) == 4
sizeof(DBL_MIN / 4.) == 8
sizeof(LDBL_MIN / 4.L) == 12

This means that FLT_MIN / 4.F only can be a float, and so on.
Moreover, FLT_MIN / 4.F must be a subnormal float number.
However, it seems like the fpclassify() macro behave as if any argument were a long double number.

Just in case, I have recompiled the program by putting the constants at hand:

printf(" Float: %x\n", fpclassify(0x1p-128F));

The result was the same.

Am I missunderstanding the C99 rules? Or the fpclassify() macro has a bug in the GCC compiler?
(in the same way, the isnormal() macro "returns" 1 for float and double, but 0 for long double).
I quote the C99 standard paragraph that explains the behaviour of fpclassify macro:

First, an argument represented in a format wider than its semantic type is converted to its semantic type.
Then classification is based on the type of the argument.

Thanks.
Sincerely, yours.
Argentinator