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我在 x86 汇编器中有一小段代码,它除以一个数字
push dword 1193182 ;push the argument mov eax, [ebp+4] push eax ;divide pop ebx pop eax div ebx push eax
在这种情况下,[ebp+4] 等于 262。但是,当我运行此代码段时,QEMU 三重故障
清除 EDX 并xor edx, edx解决问题
xor edx, edx