问题描述
谁能解释为什么这段代码会进入无限循环?
can anyone explain why this code snippet goes into an infinite loop?
我认为这与 printf 函数有关.
I presume it would have something to do with the printf function.
q1: .asciz "Hello World\n"
.global main
main:
movq %rsp, %rbp
movq $3, %rcx
jmp bottom
loop:
movq $0, %rax
movq $q1, %rdi
call printf
bottom:
decq %rcx
cmpq $0, %rcx
jne loop
end:
movq $0, %rdi
call exit
推荐答案
被调用函数需要保留的唯一寄存器是:rbp、rbx、r12、r13、r14、r15.所有其他的都可以由被调用的函数自由更改.
The only registers that the called function is required to preserve are: rbp, rbx, r12, r13, r14, r15. All others are free to be changed by the called function.
因此，很可能是 printf 正在修改 rcx 寄存器，所以它永远不会变为 0.
Therefore, the likelihood is that printf is modifying the rcx register, so it never goes to 0.
如果你推送 rcx 并稍后弹出它，那将阻止它被修改.
If you push rcx and pop it later, that would prevent it from being modified.
请注意，您似乎没有为 printf 推送参数.我认为 printf 需要 2 个参数.
Note it does not appear you are pushing args for printf. I think printf takes 2 args.
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