此英特尔jmpq指令有什么作用? [英] What does this Intel jmpq instruction do?

问题描述

如何在下面的Intel指令中计算地址0x600860? 0x4003b8 + 0x2004a2 = 60085a，所以我看不到计算是如何进行的.

How is the address 0x600860 computed in the Intel instruction below? 0x4003b8 + 0x2004a2 = 60085a, so I don't see how the computation is carried out.

0x4003b8 <puts@plt>: jmpq *0x2004a2(%rip) # 0x600860 <[e-mail]>

推荐答案

在Intel上，JMP，CALL等相对于 next 指令的程序计数器.

On Intel, JMP, CALL, etc. are relative to the program counter of the next instruction.

您遇到的下一条指令是在0x4003be和0x4003be + 0x2004a2 == 0x600860

The next instruction in your case was at 0x4003be, and 0x4003be + 0x2004a2 == 0x600860

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