新手求助

[wk8023xy]

一个96位的二进制数，分三部分分别存储在 eax， ebx， ecx。怎么转化为96位二进制对应的十进制输出？

[wk8023xy]

打不开啊

[masmaster]

[Quote=引用 2 楼 wk8023xy 的回复:]
打不开啊
[/Quote]
帮你搬过来

;conver a number result to output_buffer and fill '$' after number string
;function:
;    conver 32 bit integer result to string
;input parameter:
;     none, always convert varible result

;output parameter
;     di: The address of output buffer

output_num   proc
        push di        ;di is the address of out_buff    
    mov  bx,10    

convert_high:
    xor  dx,dx
    mov  ax,word ptr [result+2]
    div  bx
    mov  word ptr [result+2],ax
    
convert_low:
    mov  ax,word ptr [result]
    div  bx
    mov  word ptr [result],ax
    
    add  dl,'0'        ;save (result % 10)+'0' to out_buffer
    mov  byte ptr [di],dl
    inc  di            ;di point next position    

cmp_r:
    cmp  word ptr [result+2],0
    jnz  convert_high
        
        mov  dx,0
        cmp  word ptr [result],0
    jnz  convert_low
    
inv_string:    ;swap string head and string tail
    pop  si         ;now si is the address of out_buff
    dec  di        ;the di point to the last char
    mov  bx,di    
inv_loop:
        mov  al,[si]    ;swap char
        mov  ah,[di]
        mov  [di],al
        mov  [si],ah
        inc  si
        dec  di
cmp_head_tail:
    cmp  si,di
    jb   inv_loop
                
        mov  byte ptr [bx+1], '$'    ; For dos ah=09, string output, the terminal char must be '$' 
        ret
output_num   endp

[wk8023xy]

这个是32位的，96位岂不是更麻烦？

[zara]

这个，算法上可以参考http://topic.csdn.net/u/20111213/01/7b9809be-1061-4bd3-9fce-b32cadcb66e3.html这个帖子的 #4楼 里那个 output_num 子程，虽然它处理的是 2 个单元，但这个方法可以扩展到任意长度的数据。

[wk8023xy]

这个东西这么复杂啊，有点看不懂

[wk8023xy]

。。。关键是那位代码的主人不在啊，真有点头疼了！给他说话也没回，着急。。。

[zara]

这个，是有些麻烦。进制转换输出结果，一般都是采用循环除 10 得余数直至 0 再逆序输出余数（对十进制来说）。超出当前位数表示的数值的输出，就涉及到超长类数运算方面的算法或技巧了；这方面，上面那段代码的主人应该是颇为擅长的。

[liangbch]

你的开发环境是VC吗？如果是我给你写一个此文件嵌入式的汇编代码的函数。

[liangbch]

更正，“此文件嵌入式的汇编代码的函数”应该是 “c文件嵌入式的汇编代码的函数”

[liangbch]

先给出所有源代码

#include "stdio.h"
#include "stdlib.h"

typedef unsigned long DWORD;

//pn[0], the lowest 32bits
//pn[1], the middle 32bits
//pn[2], the highest 32bits

void uint96_to_string(DWORD *pn, char *szString)
{
	_asm
	{
		push esi
		push edi
		push ecx

		mov esi,pn
		mov edi,szString
		mov ecx,10
		jmp check_n

div_h_32bit:
		mov eax,dword ptr [esi+8]
		div ecx
		mov dword ptr [esi+8], eax

div_m_32bit:
		mov eax,dword ptr [esi+4]
		div ecx
		mov dword ptr [esi+4], eax

div_l_32bit:
		mov eax,dword ptr [esi]
		div ecx
		mov dword ptr [esi], eax

		add dl,'0'
		mov byte ptr [edi],dl
		inc edi

check_n:
		xor edx,edx
		cmp [esi+8],0
		jnz div_h_32bit

		cmp [esi+4],0
		jnz div_m_32bit
		
		cmp [esi],0
		jnz div_l_32bit

		//now, the whole 96bit number is full 0
		cmp edi,szString
		jne invert_n		
		
		//if uint96 is 0, then edi == szString, in this case, we need put '0' to szString and return
		mov byte ptr [edi],'0'	//put 0 to buffer 
		mov byte ptr [edi+1],0	//c language string terminating symbol
		jmp thisExit			

invert_n:
		mov byte ptr [edi],0	//c language string terminating symbol
		dec edi					//the edi point to the tail of string
		mov esi,szString		//the esi point to the 	head of string
		jmp invert_cmp

invert_loop:
		mov al,[esi]
		mov dl,[edi]
		mov [edi],al
		mov [esi],dl
		inc esi
		dec edi

invert_cmp:
		cmp esi,edi
		jb  invert_loop
thisExit:
		pop ecx
		pop edi
		pop esi
	}

}

void test_uint96_to_string()
{
	DWORD n[3];
	char result[32];
	int case_id=1;

//case 1:
	n[0]=0;
	n[1]=0;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 2:
	n[0]=9;
	n[1]=0;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 3:
	n[0]=10;
	n[1]=0;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 4:
	n[0]=0xffffffff;
	n[1]=0;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 4:
	n[0]=0;
	n[1]=1;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 6:
	n[0]=1;
	n[1]=1;
	n[2]=0;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 7:
	n[0]=0;
	n[1]=0;
	n[2]=1;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 8:
	n[0]=1;
	n[1]=0;
	n[2]=1;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;

//case 9:
	n[0]=0xffffffff;
	n[1]=0xffffffff;;
	n[2]=0xffffffff;;
	uint96_to_string(n,result);
	printf("run case %d,result is %s\n",case_id,result);
	case_id++;


}

int main(int argc, char* argv[])
{
	test_uint96_to_string();
}