问题描述
我尝试读取x86_64寄存器 rip 的值。这是 objdump 显示的内容。
I try to read the value of x86_64 register rip. Here is what objdump shows.
4017ec: 48 8d 35 00 00 00 00 lea 0x0(%rip),%rsi
4017f3: 41 89 d4 mov %edx,%r12d
我希望在执行指令 0x4017ec 之后， rsi 的值应为 0x4017ec 。但是它是 0x4017f3 ，这是下一条指令的地址。
I expect that after instruction 0x4017ec is executed, the value of rsi should be 0x4017ec. However it is 0x4017f3, which is the address of the next instruction.
我使用gdb在<$处停止c $ c> 0x4017ec ，那时 rip 的值是 0x4017ec 。为什么当时 rip 的值未加载 rsi ？处理器是否应该从 0x4017ec 中读取指令？
I use gdb to stop at 0x4017ec and at that time the value of rip is 0x4017ec. Why is rsi not loaded by the value of rip at that time? Should the processor read instruction from 0x4017ec?
推荐答案
取决于体系结构，％rip保留当前执行的指令或下一条要执行的指令。在这里，您在0x4017ec之前添加了一个断点，这意味着下一条要执行的指令是0x4017ec。但是％rsi仅在执行第一条指令后才会加载。届时％rip将已经更新为指向下一条指令。
Depends on the architecture %rip holds either the current executing instruction or the next instruction to be executed. Here you added a breakpoint before 0x4017ec which means the next instruction to be executed is 0x4017ec. But %rsi will be loaded only after executing the first instruction. By then %rip would have already updated to point to the next instruction.
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