__int64 在 32 位机器上? [英] __int64 on a 32-Bit machine?

问题描述

我刚刚在我的 32 位机器上尝试了 MSVC 2010 中的一些东西，发现我可以在我的程序中使用 __int64 - 这确实有效！

I just tried something in MSVC 2010 on my 32-bit machine here and found out that I can use __int64 in my programs - which actually work!

这怎么可能?

推荐答案

32 位算法在 16 位系统上的工作方式相同.

Same way 32-bit arithmetic worked on 16-bit systems.

在这种情况下，它使用 2 个 32 位内存地址一起组成一个 64 位数字.加法/减法很容易，您可以按部分进行，唯一的问题是将结转从较低的部分转移到较高的部分.对于乘法/除法，它更难(即更多指令).

In this case, it uses 2 32-bit memory addresses to form a 64-bit number together. Addition/substraction is easy, you do it by parts, the only gotcha is taking the carry-over from the lower part to the higher part. For multiplication/division, it's harder (ie more instructions).

它显然很慢，比乘法的 32 位算术慢很多，但如果你需要它，它就在你身边.当您升级到 64 位 处理器 编译器时，它会自动优化为具有更大字长的一条指令.

It's obviously slow, quite a bit slower than 32 bit arithmetic for multiplication, but if you need it, it's there for you. And when you upgrade to a 64-bit processor compiler, it gets automatically optimized to one instruction with the bigger word size.

Visual Studio 2010 Professional 在 32 位处理器上的 64 位乘法实现，在发布模式下编译，是:

The Visual Studio 2010 Professional implementation of 64 bit multiplication on a 32-bit processor, compiled in release mode, is:

_allmul PROC NEAR

A       EQU     [esp + 4]       ; stack address of a
B       EQU     [esp + 12]      ; stack address of b

        mov     eax,HIWORD(A)
        mov     ecx,HIWORD(B)
        or      ecx,eax         ;test for both hiwords zero.
        mov     ecx,LOWORD(B)
        jnz     short hard      ;both are zero, just mult ALO and BLO

        mov     eax,LOWORD(A)
        mul     ecx

        ret     16              ; callee restores the stack

hard:
        push    ebx

A2      EQU     [esp + 8]       ; stack address of a
B2      EQU     [esp + 16]      ; stack address of b

        mul     ecx             ;eax has AHI, ecx has BLO, so AHI * BLO
        mov     ebx,eax         ;save result

        mov     eax,LOWORD(A2)
        mul     dword ptr HIWORD(B2) ;ALO * BHI
        add     ebx,eax         ;ebx = ((ALO * BHI) + (AHI * BLO))

        mov     eax,LOWORD(A2)  ;ecx = BLO
        mul     ecx             ;so edx:eax = ALO*BLO
        add     edx,ebx         ;now edx has all the LO*HI stuff

        pop     ebx

        ret     16              ; callee restores the stack

如您所见，它比普通乘法慢很多.

As you can see, it's a LOT slower than normal multiplication.

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