x86-64 GNU 汇编
在调查崩溃时，我遇到了以下代码片段，并立即意识到 mov 指令实际上应该是 movq 以获得正确的 64 位寄存器操作。
#elif defined(__x86_64__)
unsigned long rbp;
__asm__ volatile ("mov %%rbp, %0" : "=r" (rbp));
sp = (void **) rbp;
#else
除此之外，我还发现文档声称 x86-64 的 rbp 寄存器是通用的，不包含当前帧的地址。我还发现文档声称 rbp 确实包含当前帧的地址。有人可以澄清一下吗？
While investigating a crash, I came across the following code snippet and immediately recognized that the mov instruction should actually be movq to get the correct 64-bit register operation.
#elif defined(__x86_64__)
unsigned long rbp;
__asm__ volatile ("mov %%rbp, %0" : "=r" (rbp));
sp = (void **) rbp;
#else
Further to this, I also found documentation that claims that the rbp register for x86-64 is general purpose and does not contain the address of the current frame. I have also found documentation that claims that rbp does contain the address of the current frame. Can someone clarify?
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关于问题的第一部分（
movq而不是mov），汇编器（在本例中）将识别出您的操作数是 64 位，并且会正确地使用movq。mov不是一条有效的指令，它是一种告诉汇编器“根据操作数使用正确的mov变体”的方法。关于第二部分，它实际上是两者：它是一个通用寄存器，从某种意义上说，它可以保存任何值。它还用作堆栈帧基指针。 AMD64 应用程序编程手册的“2.4 堆栈操作”部分说：
Regarding the first part of your question (
movqinstead ofmov), the assembler (as, in this case), will recognize that your operand is 64 bits, and will correctly usemovq.movis not a valid instruction, it's a way to tell the assembler "use the rightmovvariant depending on the operands".Regarding the second part, it's actually both: it's a general purpose register, in the sense that it can hold any value. It is also used as a stack-frame base pointer. The '2.4 Stack operation' section of the AMD64 Application programming manual says: