使用 %cl 左移 [英] Left shift using %cl

问题描述

假设我有一个二进制数.1010 是十进制的 10.

Let's say I have a number in binary. 1010 which is 10 in decimal.

我知道左移 1 位本质上是将数字乘以 2.

I understand that shifting left by 1 bit is essentially multiplying the number by 2.

教科书中的一句话让我感到困惑.

Theres a line in a textbook that's got me confused.

salq %cl, %rdx

%rdx 是一个数字，%salq 是一个左移.我感到困惑的是 %cl.

%rdx is a number and %salq is a left shift. What I'm confused about is the %cl.

我读过 CL 是 8 位，这是否意味着我要乘以 2^8?

I've read that CL is 8 bits, does that mean I'm multiplying by 2^8?

推荐答案

cl 部分寄存器(实际上是寄存器 rcx 的最低 8 位)包含rdx 将左移的值.它有 8 位长，但移位的量是实际存在的:

The cl partial register (which is really the lowest 8 bits of the register rcx) contains the value by which rdx will be shifted left. It's eight bits long, but the amount shifted is whatever is actually in there:

movb $10, %cl
salq %cl, %rdx ; rdx is shifted 10 bits left.

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