【汇编程序】实现输出1000以内所有素数

程序需求：编程写一个完整的程序,求出1000以内的所有素数，并将它们存入Prime数组中，素数的个数存入变量Pcounter中。
编程思路：esi相当与C中的i，edi相当与C中的j，ecx相当于C中的Pcounter。

开发环境

Win10 + VS2017

C语言代码实现如下：

#include <stdio.h>
int Prime[1000];
int main()
{
	int Pcounter = 0;
	for (int i = 6; i < 1000; i++)
	{
		for (int j = 2; j < i/2; j++)
		{
			if (i%j == 0)
				break;
			if (j == i / 2 - 1)
				Prime[Pcounter++] = i;
		}
	}
	printf("2\t3\t5\t");
	for (int i = 0; i < Pcounter; i++)
		printf("%d\t", Prime[i]);
	return 0;
}

汇编语言代码实现如下：

INCLUDELIB kernel32.lib
INCLUDELIB ucrt.lib
INCLUDELIB legacy_stdio_definitions.lib

.386
.model flat,stdcall

ExitProcess PROTO,
dwExitCode:DWORD

printf    PROTO C : dword,:vararg
scanf    PROTO C : dword,:vararg

.data
format byte '%d',9,0;
msg byte 32h,9,33h,9,35h,9,0
Prime dword 0 dup(1000)

.code
main Proc
	xor ecx,ecx;ecx==Pcounter
	mov esi,6;esi==i
	jmp testing
body:
	mov edi,2;cdi==j
	jmp testing2
body2:
	mov eax,esi
	mov ebx,edi
	cdq
	div ebx
	cmp edx,0
	jne next
	jmp over
next:
	mov eax,esi
	mov ebx,2
	cdq
	div ebx
	dec eax
	cmp edi,eax
	jne over2
	mov dword ptr Prime[ecx*4],esi
	inc ecx
over2:
	inc edi
testing2:
	mov eax,esi
	mov ebx,2
	cdq
	div ebx
	cmp edi,eax
	jl body2

over:
	inc esi
testing:
	cmp esi,1000
	jl body

	pushad
	invoke printf,offset msg
	popad

	xor esi,esi
again:
	pushad
	invoke printf,offset format,dword ptr Prime[esi*4]
	popad
	inc esi
	loop again

	push 0h
	call ExitProcess
main endp
end main

编译运行后结果如下：