x86组装中的力量 [英] power of in x86 assembly

问题描述

作为ASM编程的入门者，我需要将2的结果转换为Assembly的38的幂，并且我需要您的帮助以了解为什么我的程序未产生我需要的结果(它打印4十进制):

as a starter in ASM programming i am need to get the result of 2 to the power of 38 in Assembly , and i need your help in understanding why is my program doesn't produce the result i am need (it prints 4 decimal):

.386
.model flat, stdcall
option casemap:none

include \masm32\include\windows.inc
include \masm32\include\msvcrt.inc
includelib \masm32\lib\msvcrt.lib

.data

formatstr db "%d",0

.code 
start:

mov eax , 2
mov ecx , 38
mov esi , eax
mov edx , 0

.while TRUE
    mul esi
    mov esi, edx
    add esi, eax
    mov edx, 0
    mov eax, 2
    dec ecx
    .break .if (!ecx)
.endw




invoke crt__cprintf, addr formatstr, esi


end start

如您所见，我正在使用masm32编写代码(如果在这种情况下有问题的话).

as you can see i am writing it using masm32 (if that has any matter in that case).

THX.

推荐答案

2^38显然不适合一个32位寄存器，例如eax.

2^38 obviously does not fit in one 32-bit register such as eax.

要存储值2^38(274877906944)，您需要39位.在32位代码中，您可以使用例如.两个32位寄存器，例如edx:eax.但是，在32位代码中，mul仅接受32位因数(例如，寄存器，其他始终为eax)，因此在循环中使用32位mul可能会获胜不起作用，因为即使mul将64位结果存储在 的32位寄存器中>.

To store the value 2^38 (274877906944) you need 39 bits. In 32-bit code you can use eg. two 32-bit registers such as edx:eax. However, in 32-bit code mul only accepts 32-bit factors (eg. registers, other of them is always eax), so using 32-bit mul in a loop won't work, because you cannot store your intermediate results in a 32-bit register to be multiplied again, even if mul stores the 64-bit result in edx:eax.

但是您可以使用 rcl 来计算2^38以32位代码显示:

But you can use rcl to compute eg. 2^38 in 32-bit code:

    xor edx,edx
    mov eax,2    ; now you have 2 in edx:eax
    mov ecx,38   ; 2^n, in this case 2^38 (any value x, 1 <= x <= 63, is valid).

x1: dec ecx      ; decrease ecx by 1
    jz ready     ; if it's 2^1, we are ready.

    shl eax,1    ; shift eax left through carry flag (CF) (overflow makes edx:eax zero)
    rcl edx,1    ; rotate edx through carry flag (CF) left
    jmp x1

ready:            ; edx:eax contains now 2^38.

编辑:一个受@Jagged O'Neill的答案启发的非循环实现.对于指数> = 32，这一次没有跳跃，对于指数<<是1，没有跳跃. 32，也适用于ecx 0，大于63的ecx设置edx:eax至0.

a non-loop implementation inspired by @Jagged O'Neill's answer. This one is without jumps for exponent >= 32, one jump for exponent < 32, works also for ecx 0, for ecx greater than 63 sets edx:eax to 0.

    mov     ecx,38          ; input (exponent) in ecx. 2^n, in this case 2^38.
                            ; (any value x, 0 <= x <= 63, is valid).
; the code begins here.

    xor     eax,eax
    xor     edx,edx         ; edx:eax is now prepared.

    cmp     cl,64           ; if (cl >= 64),
    setb    al              ; then set eax to 0, else set eax to 1.
    jae     ready           ; this is to handle cl >= 64.

; now we have 0 <= cl <= 63

    sub     ecx,1
    setnc   al              ; if (count == 0) then eax = 0, else eax = 1.
    lea     eax,[eax+1]     ; eax = eax + 1. does not modify any flags.
    jna     ready           ; 2^0 is 1, 2^1 = 2, those are ready now.
    mov     ebx,ecx         ; copy ecx to ebx
    cmp     cl,32           ; if (cl >= 32)
    jb      low_5_bits
    mov     cl,31           ; then shift first 31 bits to the left.
    shld    edx,eax,cl
    shl     eax,cl          ; now shifted 31 bits to the left.
    lea     ecx,[ebx-31]    ; cl = bl - 31

low_5_bits:
    shld    edx,eax,cl
    shl     eax,cl

ready:

这篇关于x86组装中的力量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！