_fto132proc bug

.cpp

#include <fenv.h>
#include <stdint.h>
#include <stdio.h>

extern "C" int64_t fto132proc(/* Passing by ST(0) */);

int main()
{
	double f = UINT32_MAX + 0.9;
	uint32_t h, l;
	//printf("%x\n", fegetround());
	//fesetround(f > 0 ? FE_UPWARD : FE_DOWNWARD);
#ifdef _MSC_VER
	__asm
	{
		fld f
		call fto132proc
		mov h, edx
		mov l, eax
	}
#else
	asm(
		"fldl %2 \n\t"
		"call fto132proc \n\t"
		"movl %%edx, %0 \n\t"
		"movl %%eax, %1 \n\t"
		: "=d" (h), "=a" (l)
		: "m"(f)
		: "st"
	);
#endif
	printf("%lld\n", h * 1ULL << 32 | l);
	printf("%llx\n", h * 1ULL << 32 | l);
}

.md

A bug in Intel® 64 and IA-32 architectures optimization reference manual.

A buggy example code is in 3.8.4.1 Rounding Mode.

add ecx,7fffffffh ; If diff<0 then decrement integer
adc eax,0 ; INC EAX (add CARRY flag)
ret

add ecx, 7fffffffh ; If diff<0 then decrement integer
sbb eax, 0 ; DEC EAX (subtract CARRY flag)
ret

In 32-bit x86 ABI, EDX and EAX registers are used to represent 64-bit integers.
EDX used for upper 32-bit value, EAX used for lower 32-bit value.
For example, the value 0x1234567890 is EDX = 0x12, EAX = 0x34567890.
If carry or borrow occurred in EAX, it must be reflected in EDX.
But the above code forgets it.
Therefore subtracting 1 from 0x100000000(EDX = 1, EAX = 0) will be 0x1FFFFFFFF(EDX = 1, EAX = 0xFFFFFFFF).
Adding 1 to -1 (EDX = 0xFFFFFFFF, EAX = 0xFFFFFFFF) will be -4294967296 (EDX = 0xFFFFFFFF, EAX = 0).

_ftol2 of VC ++ refers to Intel code.
But VC++'s _ftol2 is fixed bug.