超简单的链表反汇编分析

for(int i=0;i<20;++i){0041155E  mov         dword ptr [i],0 //初始化i00411565  jmp         wmain+30h (411570h) //以下是++i00411567  mov         eax,dword ptr [i] 0041156A  add         eax,1 //++i0041156D  mov         dword ptr [i],eax //给局部变量00411570  cmp         dword ptr [i],0//比较i与2000411574  jge         wmain+53h (411593h) //如果大于就跳出去
    printf("%dn",i);00411576  mov         esi,esp 00411578  mov         eax,dword ptr [i] 0041157B  push        eax  0041157C  push        offset string "%dn" (415644h) 00411581  call        dword ptr [__imp__printf (4182CCh)] //printf打印00411587  add         esp,8 //堆栈平衡
0041158A  cmp         esi,esp //安全检查0041158C  call        @ILT+325(__RTC_CheckEsp) (41114Ah) 
  } 00411591  jmp         wmain+27h (411567h) //执行完跳转到++i

#include "stdafx.h"#include <stdlib.h>struct node{  int x;  struct node * next;};typedef struct node pnode;int main(int argc, char* argv[]){  pnode* head = (pnode*)malloc(sizeof(node));  pnode* curr;  int i;    for(i=1;i<=10;i++){    curr=(pnode*)malloc(sizeof(pnode));    curr->x = i;    curr->next = NULL;
    head->next = curr;//下一个节点指向head    head = head->next;//本节点给head  }

  while(head){    printf("%d",head->x);    curr->next;  }  return 0;}