技術標籤：逆戰長征c語言

題目：將一個數組逆序輸出。

程式碼（當然還可以利用C++中堆疊方法等，這裡就不擴充套件了關鍵看看編譯器怎麼寫的彙編）

#include<stdio.h>
#define N 5
int main()
{
	int a[N]={3,5,9,1,6},i,temp;
	printf("原始資料為:\n");
	for(i=0;i<N;i++)
		printf("%4d",a[i]);
	for(i=0;i<N/2;i++)
	{
		temp=a[i];
		a[i]=a[N-i-1];
		a[N-i-
 
1]=temp;
	}
	printf("\n倒序後資料為:\n");
	for(i=0;i<N;i++)
		printf("%4d",a[i]);
	printf("\n");
	return 0;
}

顯示

彙編

3: int main()
4: {
0040D7C0 push ebp
0040D7C1 mov ebp,esp
0040D7C3 sub esp,5Ch
0040D7C6 push ebx
0040D7C7 push esi
0040D7C8 push edi
0040D7C9 lea edi,[ebp-5Ch]
0040D7CC mov ecx,17h
0040D7D1 mov eax,0CCCCCCCCh

0040D7D6 rep stos dword ptr [edi]
5: int a[N]={3,5,9,1,6},i,temp;
0040D7D8 mov dword ptr [ebp-14h],3
0040D7DF mov dword ptr [ebp-10h],5
0040D7E6 mov dword ptr [ebp-0Ch],9
0040D7ED mov dword ptr [ebp-8],1
0040D7F4 mov dword ptr [ebp-4],6
6: printf(“原始資料為:\n”);
0040D7FB push offset string “\xd4\xad\xca\xbc\xca\xfd\xbe\xdd\xce\xaa:\n” (00422fb4)
0040D800 call printf (00401130)
0040D805 add esp,4
7: for(i=0;i<N;i++)
0040D808 mov dword ptr [ebp-18h],0
0040D80F jmp main+5Ah (0040d81a)
0040D811 mov eax,dword ptr [ebp-18h]
0040D814 add eax,1
0040D817 mov dword ptr [ebp-18h],eax
0040D81A cmp dword ptr [ebp-18h],5
0040D81E jge main+77h (0040d837)
8: printf("%4d",a[i]);
0040D820 mov ecx,dword ptr [ebp-18h]
0040D823 mov edx,dword ptr [ebp+ecx4-14h]
0040D827 push edx
0040D828 push offset string “ret=%d” (0042201c)
0040D82D call printf (00401130)
0040D832 add esp,8
0040D835 jmp main+51h (0040d811)
9: for(i=0;i<N/2;i++)
0040D837 mov dword ptr [ebp-18h],0
0040D83E jmp main+89h (0040d849)
0040D840 mov eax,dword ptr [ebp-18h]
0040D843 add eax,1
0040D846 mov dword ptr [ebp-18h],eax
0040D849 cmp dword ptr [ebp-18h],2
0040D84D jge main+0BDh (0040d87d)
10: {
11: temp=a[i];
0040D84F mov ecx,dword ptr [ebp-18h]
0040D852 mov edx,dword ptr [ebp+ecx4-14h]
0040D856 mov dword ptr [ebp-1Ch],edx
12: a[i]=a[N-i-1];
0040D859 mov eax,5
0040D85E sub eax,dword ptr [ebp-18h]
0040D861 mov ecx,dword ptr [ebp-18h]
0040D864 mov edx,dword ptr [ebp+eax4-18h]
0040D868 mov dword ptr [ebp+ecx4-14h],edx
13: a[N-i-1]=temp;
0040D86C mov eax,5
0040D871 sub eax,dword ptr [ebp-18h]
0040D874 mov ecx,dword ptr [ebp-1Ch]
0040D877 mov dword ptr [ebp+eax4-18h],ecx
14: }
0040D87B jmp main+80h (0040d840)
15: printf("\n倒序後資料為:\n");
0040D87D push offset string “\n\xb5\xb9\xd0\xf2\xba\xf3\xca\xfd\xbe\xdd\xce\xaa:\n” (00422fa4)
0040D882 call printf (00401130)
0040D887 add esp,4
16: for(i=0;i<N;i++)
0040D88A mov dword ptr [ebp-18h],0
0040D891 jmp main+0DCh (0040d89c)
0040D893 mov edx,dword ptr [ebp-18h]
0040D896 add edx,1
0040D899 mov dword ptr [ebp-18h],edx
0040D89C cmp dword ptr [ebp-18h],5
0040D8A0 jge main+0F9h (0040d8b9)
17: printf("%4d",a[i]);
0040D8A2 mov eax,dword ptr [ebp-18h]
0040D8A5 mov ecx,dword ptr [ebp+eax4-14h]
0040D8A9 push ecx
0040D8AA push offset string “ret=%d” (0042201c)
0040D8AF call printf (00401130)
0040D8B4 add esp,8
0040D8B7 jmp main+0D3h (0040d893)
18: printf("\n");
0040D8B9 push offset string “\n” (00422234)
0040D8BE call printf (00401130)
0040D8C3 add esp,4
19: return 0;
0040D8C6 xor eax,eax
20: }
0040D8C8 pop edi
0040D8C9 pop esi
0040D8CA pop ebx
0040D8CB add esp,5Ch
0040D8CE cmp ebp,esp
0040D8D0 call __chkesp (004011b0)
0040D8D5 mov esp,ebp
0040D8D7 pop ebp
0040D8D8 ret

2021/2/14