什么是LEA指令的目的是什么？ [英] What's the purpose of the LEA instruction?

问题描述

对于我来说，这似乎只是一个时髦的MOV。什么是它的目的，当我应该使用它？

For me, it just seems like a funky MOV. What's its purpose and when should I use it?

推荐答案

正如其他人所指出的那样，LEA（装载有效地址）经常被用来作为一个绝招做某些计算，但这并不是它的主要目的。 x86指令集被设计为支持高级语言如Pascal和C，其中阵列与MDASH;特别是整数或小的结构＆MDASH的阵列，是常见的。举个例子，一个结构重新presenting（X，Y）坐标：

As others have pointed out, LEA (load effective address) is often used as a "trick" to do certain computations, but that's not its primary purpose. The x86 instruction set was designed to support high-level languages like Pascal and C, where arrays—especially arrays of ints or small structs—are common. Consider, for example, a struct representing (x, y) coordinates:

struct Point
{
     int xcoord;
     int ycoord;
};

现在想像这样一种说法：

Now imagine a statement like:

int y = points[i].ycoord;

其中，点[] 是点的数组。假设数组的基本已在 EBX 和变量 I 是 EAX 和 XCOORD 和 YCOORD 均为32位（因此 YCOORD 以抵消结构4个字节），这句话可以被编译为：

where points[] is an array of Point. Assuming the base of the array is already in EBX, and variable i is in EAX, and xcoord and ycoord are each 32 bits (so ycoord is at offset 4 bytes in the struct), this statement can be compiled to:

MOV EDX, [EBX + 8*EAX + 4]    ; right side is "effective address"

将登陆是在 EDX 。 8的比例因子是因为每个点的大小8个字节。现在考虑与运营商和放大器地址所用的相同的前pression;

which will land y in EDX. The scale factor of 8 is because each Point is 8 bytes in size. Now consider the same expression used with the "address of" operator &:

int *p = &points[i].ycoord;

在这种情况下，你不希望 YCOORD 的价值，但它的地址。这就是 LEA （加载有效地址）的用武之地。取而代之的是 MOV ，编译器可以生成

In this case, you don't want the value of ycoord, but its address. That's where LEA (load effective address) comes in. Instead of a MOV, the compiler can generate

LEA ESI, [EBX + 8*EAX + 4]

将在 ESI 加载地址。

这篇关于什么是LEA指令的目的是什么？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！