摘要：
int main(int argc, char* argv[]){ int i; scanf("%d", &i); int j = i / 2; printf("%d", j);return 0;}编译后：10: int j = i / 2;00401039 mov eax,dword ptr [ebp-4]0040103C cdq0040103D sub eax,edx0040103F sar eax,100401041 mov dword ptr [ebp-8],eax即：M / 2 为mov eax,Mcdqsub eax,edxsar e 阅读全文
