#1927 fpclassify macro and subnormal numbers

[argentinator]

I think that probably there is a bug with the macro fpclassify(). The following
program classifies a 'double subnormal' number as a 'normal' one (maybe the
value it is seen by the macro as a 'long double').

---

#include <math.h>
#include <float.h>
#include <stdio.h>
int main(void) {
    double x = DBL_MIN/1024.0;
    long double z = LDBL_MIN/1024.0;
    printf("x == %a\n\nclass x == %X\n\nclass z == %X\n", 
           x, fpclassify(x), fpclassify(z));
}

---

My compiler is GCC 4.7.2 on MinGW, under Windows 7 (running just in CMD.EXE).
The compiler option is just: -std=c99. (Anyway, without command line options I
have found the same results).
I have FLT_EVAL_METHOD == 2.

The output I have obtained is:

---

x == 0x8p-1035

class x == 400

class z == 4400

---

Comments:

• x is a 'double subnormal' number.
• fpclassify(x) tells that x is a 'normal' value (0x400), and this would be
  wrong.
• fpclassify(z) tells that z is a 'long double subnormal' value (0x4400), which
  is right.

It seems that fpclassify(x) have not recognized that x has type 'double'.

[argentinator]

The source code is not properly shown in the 1st message,
so I add an atachment.

[Keith Marshall]

• Description has changed:

Diff:

--- old
+++ new
@@ -3,6 +3,7 @@
 value it is seen by the macro as a 'long double').

 ----------------------------------------------------------------
+~~~~
 #include <math.h>
 #include <float.h>
 #include <stdio.h>
@@ -12,6 +13,7 @@
     printf("x == %a\n\nclass x == %X\n\nclass z == %X\n", 
            x, fpclassify(x), fpclassify(z));
 }
+~~~~
 ----------------------------------------------------------------

 My compiler is GCC 4.7.2 on MinGW, under Windows 7 (running just in CMD.EXE).
@@ -22,11 +24,13 @@
 The output I have obtained is:

 --------------------------------------------------------------
+~~~~
 x == 0x8p-1035

 class x == 400

 class z == 4400
+~~~~
 --------------------------------------------------------------

 Comments:

[argentinator]

Keith, thank you for fixing my original message.

[Earnie Boyd]

I'll have to take a look at the implementation. I'm planning a 4.0 release soon which is why I've marked this for 4.1.