jmp指令在这种情况下如何在att程序集中工作 [英] How does the jmp instruction work in att assembly in this instance

问题描述

我现在正在逐步执行 AT&T 程序集中的一个函数，但无法弄清楚这个特定的 jmp 命令是如何工作的.

I am stepping through a function in AT&T assembly right now and can't figure out how this specific jmp command works.

    jmp    *0x804a140(,%eax,4)

它究竟是如何使用 %eax 寄存器和 4 与跳转指令的?我以前从未见过 jmp 用这种方式.

How exactly is it using the %eax register and 4 with the jump instruction? I have never seen jmp used this way before.

推荐答案

如果您对 at&t 语法感到困惑，请将您的工具切换到 intel 模式.

If you are confused by at&t syntax, switch your tool to intel mode.

您看到的有效地址并非特定于跳转，您可能会在任何需要内存操作数的指令中遇到它.

The effective address you see is not specific to jumps, you could have encountered it with any instruction that takes a memory operand.

在英特尔语法中，这看起来像:jmp [0x804a140 + 4 * eax].这是一个间接跳转，从内存地址 0x804a140 + 4 * eax 获取跳转目标.这可能是所谓的跳转表中的一项.

In intel syntax this would look like: jmp [0x804a140 + 4 * eax]. It's an indirect jump that fetches the jump target from memory address 0x804a140 + 4 * eax. This is probably an item in a so-called jump table.

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