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It appears that microsoft decided NAN will be represented by '\x00\x00\x00\x00\x00\x00\xf8\xff', which has the sign bit set. 
Compiling this c code with visual 9.0 gives the correct answers for the first value, and a mess for the second:
#include <math.h>
#include <stdio.h>
#include <float.h>
int main( void ) {
 unsigned long nan[2]={0xffffffff, 0x7fffffff};
 double g;
 double z, zn;
 int i;
 for (i=0;i<2; i++)
 { 
 g = *( double* )(nan+i);
 printf( "g( %g ) is NaN, _isnan(g) %d\n", g, _isnan(g) );
 z = _copysign(-3, g);
 zn = _copysign(-3, -g);
 printf("z=%g, zn=%g\n", z, zn);
 }
 return 0;
}
This corresponds with loewis 's observation.