了解ATT汇编语言 [英] Understanding ATT Assembly Language

问题描述

C版本:

int arith(int x, int y, int z)
{
    int t1 = x+y;
    int t2 = z*48;
    int t3 = t1 & 0xFFFF;
    int t4 = t2 * t3;
    return t4;
}

同一程序的ATT汇编版本:

ATT Assembly version of the same program:

x在％ebp + 8，y在％ebp + 12，z在％ebp + 16

x at %ebp+8, y at %ebp+12, z at %ebp+16

movl   16(ebp), %eax    
leal   (%eax, %eax, 2), %eax   
sall   $4, %eax      // t2 = z* 48... This is where I get confused
movl   12(%ebp), %edx   
addl   8(%ebp), %edx
andl   $65535, %edx
imull  %edx, %eax

除了左移，我了解它在程序的所有点上所做的一切.

I understand everything it is doing at all points of the program besides the shift left.

我认为它将向左移动4次.为什么会这样?

I assume it is going to shift left 4 times. Why is that?

谢谢！

我也理解我困惑的部分等同于C版本的z * 48部分.

I also understand that the part I'm confused on is equivalent to the z*48 part of the C version.

我不了解的是左移4次等于z * 48.

What I'm not understanding is how does shifting left 4 times equate to z*48.

推荐答案

您错过了leal (%eax, %eax, 2), %eax行.应用一些数学，汇编代码将读取:

You missed the leal (%eax, %eax, 2), %eax line. Applying some maths the assembly code reads:

a := x
a := a + 2*a    // a = 3*x
a := a * 2^4    // a = x * 3*16

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