汇编实现(AT&T语法):
.section .data values: .int 33, 25, 67, 10, 1 .section .text .global _start _start: nop movl $values, %esi movl $4, %ecx movl $4, %ebx loop: movl (%esi), %eax cmp %eax, 4(%esi) jge skip xchg %eax, 4(%esi) movl %eax, (%esi) skip: add $4, %esi dec %ebx jnz loop dec %ecx jz end movl $values, %esi movl %ecx, %ebx jmp loop end: movl $1, %eax movl $0, %ebx int $0x80
C 实现:
#include#include void PrintValues(int *values, int count) { if (NULL == values || 0 >= count) return; for (int i = 0; i < count; ++i) { printf("%d,", *(values+i)); } printf("/n"); } int main(int argc, char** argv) { int values[] = {33, 25, 67, 10, 1}; int count = sizeof(values) / sizeof(*values); PrintValues(values, count); for (int outer = 0; outer < count-1; ++outer) { for (int inner = 0; inner < count-1-outer; ++inner) { if (*(values+inner) > *(values+inner+1)) { int temp = *(values+inner); *(values+inner) = *(values+inner+1); *(values+inner+1) = temp; } else { continue; } } } PrintValues(values, count); getchar(); return 0; }
C 实现的反汇编:
int main(int argc, char** argv) { push ebp mov ebp,esp sub esp,110h push ebx push esi push edi lea edi,[ebp-110h] mov ecx,44h mov eax,0CCCCCCCCh rep stos dword ptr es:[edi] mov eax,dword ptr [___security_cookie (257000h)] xor eax,ebp mov dword ptr [ebp-4],eax int values[] = {33, 25, 67, 10, 1}; mov dword ptr [ebp-1Ch],21h mov dword ptr [ebp-18h],19h mov dword ptr [ebp-14h],43h mov dword ptr [ebp-10h],0Ah mov dword ptr [ebp-0Ch],1 int count = sizeof(values) / sizeof(*values); mov dword ptr [ebp-28h],5 PrintValues(values, count); mov eax,dword ptr [ebp-28h] push eax lea ecx,[ebp-1Ch] push ecx call PrintValues (2511BDh) add esp,8 for (int outer = 0; outer < count-1; ++outer) mov dword ptr [outer],0 jmp main+74h (253604h) mov eax,dword ptr [outer] add eax,1 mov dword ptr [outer],eax mov eax,dword ptr [ebp-28h] sub eax,1 cmp dword ptr [outer],eax jge main+0D9h (253669h) { for (int inner = 0; inner < count-1-outer; ++inner) mov dword ptr [inner],0 jmp main+91h (253621h) mov eax,dword ptr [inner] add eax,1 mov dword ptr [inner],eax mov eax,dword ptr [ebp-28h] sub eax,1 sub eax,dword ptr [outer] cmp dword ptr [inner],eax jge main+0D7h (253667h) { if (*(values+inner) > *(values+inner+1)) mov eax,dword ptr [inner] mov ecx,dword ptr [inner] mov edx,dword ptr [ebp+eax*4-1Ch] cmp edx,dword ptr [ebp+ecx*4-18h] jle main+0D3h (253663h) { int temp = *(values+inner); mov eax,dword ptr [inner] mov ecx,dword ptr [ebp+eax*4-1Ch] mov dword ptr [temp],ecx *(values+inner) = *(values+inner+1); mov eax,dword ptr [inner] mov ecx,dword ptr [inner] mov edx,dword ptr [ebp+ecx*4-18h] mov dword ptr [ebp+eax*4-1Ch],edx *(values+inner+1) = temp; mov eax,dword ptr [inner] mov ecx,dword ptr [temp] mov dword ptr [ebp+eax*4-18h],ecx } else jmp main+0D5h (253665h) { continue; jmp main+88h (253618h) } } jmp main+88h (253618h) } jmp main+6Bh (2535FBh) PrintValues(values, count); mov eax,dword ptr [ebp-28h] push eax lea ecx,[ebp-1Ch] push ecx call PrintValues (2511BDh) add esp,8 getchar(); mov esi,esp call dword ptr [__imp__getchar (2582B0h)] cmp esi,esp call @ILT+295(__RTC_CheckEsp) (25112Ch) return 0; xor eax,eax }
在用 C 实现时,思考的核心是:如何控制循环,并在正确的时候交换数据。
在用汇编实现时,思考的核心是:如何分配寄存器,如何具体控制循环(loop 与 skip)。
在用汇编时,我先是思考好程序的流程:loop 与 skip,然后建立好对应的结构,再往结构中填写代码,然后控制好程序的执行流程。
汇编比 C 需要思考更多的问题,对程序员的要求也相对高点,C 不需要思考的问题,汇编都需要思考。在用 C# 与 Java 实现冒泡排序时,相对于C,不再需要思考指针操作相关的问题了。当用 Python 等动态语言实现这个算法时,相对于 Java 与 C#,不再需要交换数据时的临时变量,代码又节省了。