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    • [高分急等]汇编语言，请大神修改代码....

    密码设置模拟
    要求第一次能设置密码并保存，然后可以修改密码。。。不知道程序错在哪，请大神修改完善功能....
    DATA SEGMENT
    FILE DB 'E:\PW.BAT',00
    NPW DB 'Please enter your password and press the return key:',13,10,36
    INFO1 DB 0DH,0AH,'--------------------------',0DH,0AH,'You can make two choice:',0DH,0AH,0DH,0AH,'$'
    INFO2 DB '1.Reset passport',0DH,0AH,'$'
    INFO3 DB '2.Exit',0DH,0AH,'$'
    INFO4 DB 0DH,0AH,'-------------------',0DH,0AH,'|INVALID SELECTION|',0DH,0AH,'-------------------',0DH,0AH,'$'
    OPW DB 'Please input your old password and press the return key:',13,10,36
    ERR DB 13,10,'YOUR PASSWORD ERROR!',36
    MSG DB 13,10,'INPUT NEW PASSWORD:',36
    PW DB 19,18 DUP(?)
    PWS DB 19 DUP(?)

    DATA ENDS
    CODE SEGMENT
    ASSUME CS:CODE,DS:DATA,ES:DATA
    START:
    MAIN PROC NEAR
    MOV AX,DATA
    MOV DS,AX
    MOV ES,AX

    LEA DX,FILE
    MOV AX,3D02H
    INT 21H
    JC L1
    JMP HK
    L1:
    LEA DX,NPW
    MOV AH,09H
    INT 21H
    CALL SET
    LEA DX,FILE
    MOV CX,00H
    MOV AH,3CH
    INT 21H
    MOV BX,AX
    CALL SET1
    JMP HK
    HK:
    LEA DX,INFO1
    MOV AH,09H
    INT 21H
    LEA DX,INFO2
    INT 21H
    LEA DX,INFO3
    INT 21H
    MOV AH,01H
    INT 21H
    CMP AL,'1'
    JE LA
    CMP AL,'2'
    JE EXIT
    LEA DX,INFO4
    MOV AH,09H
    INT 21H
    JMP HK

    LA:
    PUSH AX
    LEA DX,OPW
    MOV AH,09H
    INT 21H
    CALL SET
    POP BX
    PUSH BX
    LEA DX,PWS
    MOV CX,1
    MOV AH,3FH
    INT 21H
    MOV CX,WORD PTR PWS
    DEC CX
    LEA DX,PWS+1
    MOV AH,3FH
    INT 21H
    LEA SI,PW
    LEA DI,PWS
    CLD
    MOV CL,PW
    XOR CH,CH
    REPE CMPSB
    CMP CX,0
    JZ LD
    LC:
    LEA DX,ERR
    MOV AH,09H
    INT 21H
    JMP HK
    LD:
    LEA DX,MSG
    MOV AH,09H
    INT 21H
    CALL SET
    POP BX
    MOV AH,3EH
    INT 21H
    LEA DX,FILE
    MOV AX,3D01H
    INT 21H
    CALL SET1

    EXIT:
    MOV AH,4CH
    INT 21H
    MAIN ENDP

    SET PROC NEAR
    MOV CL,12H
    CLD
    LEA SI,PW
    MOV SI,13H
    INC SI
    MOV DL,'*'
    I1:
    MOV AH,08H
    INT 21H
    CMP AL,0DH
    JZ R1
    MOV AH,02H
    CMP CL,0
    JNZ I2
    MOV DL,07H
    INT 21H
    JMP I1

    I2:
    MOV [SI],AL
    INT 21H
    INC SI
    DEC CL
    JMP I1
    R1:
    SUB PW,CL
    RET
    SET ENDP

    SET1 PROC NEAR
    LEA DX,PW
    MOV CL,PW
    AND CX,0FFH
    MOV AH,40H
    INT 21H
    RET
    SET1 ENDP

    CODE ENDS
    END START

    推荐答案   2011-06-29
    DATA SEGMENT
    FILE DB 'E:\PW.BAT',00
    NPW DB 'Please enter your password and press the return key:',13,10,36
    INFO1 DB 0DH,0AH,'--------------------------',0DH,0AH,'You can make two choice:',0DH,0AH,0DH,0AH,'$'
    INFO2 DB '1.Reset passport',0DH,0AH,'$'
    INFO3 DB '2.Exit',0DH,0AH,'$'
    INFO4 DB 0DH,0AH,'-------------------',0DH,0AH,'|INVALID SELECTION|',0DH,0AH,'-------------------',0DH,0AH,'$'
    OPW DB 'Please input your old password and press the return key:',13,10,36
    ERR DB 13,10,'YOUR PASSWORD ERROR!',36
    MSG DB 13,10,'INPUT NEW PASSWORD:',36
    PW DB 19,18 DUP(?)
    PWS DB 19 DUP(?)

    DATA ENDS
    CODE SEGMENT
    ASSUME CS:CODE,DS:DATA,ES:DATA
    START:
    MAIN PROC NEAR
    MOV AX,DATA
    MOV DS,AX
    MOV ES,AX

    LEA DX,FILE
    MOV AX,3D02H
    INT 21H
    JC L1
    mov bx,ax ;保存文件句柄
    JMP HK
    L1:
    LEA DX,NPW
    MOV AH,09H
    INT 21H
    CALL SET
    LEA DX,FILE
    MOV CX,00H
    MOV AH,3CH
    INT 21H
    MOV BX,AX
    CALL SET1
    JMP HK
    HK:
    LEA DX,INFO1
    MOV AH,09H
    INT 21H
    LEA DX,INFO2
    INT 21H
    LEA DX,INFO3
    INT 21H
    MOV AH,01H
    INT 21H
    CMP AL,'1'
    JE LA
    CMP AL,'2'
    JE EXIT
    LEA DX,INFO4
    MOV AH,09H
    INT 21H
    JMP HK

    LA:
    ;PUSH AX 用bx保存句柄，直接用就可以，没必要用堆栈了
    LEA DX,OPW
    MOV AH,09H
    INT 21H
    CALL SET
    LEA DX,FILE
    MOV AX,3D01H
    INT 21H
    MOV BX,AX
    LEA DX,PWS
    MOV CX,1
    MOV AH,3FH
    INT 21H
    MOV CX,WORD PTR PWS
    DEC CX
    LEA DX,PWS+1
    MOV AH,3FH
    INT 21H

    LEA SI,PW
    LEA DI,PWS
    CLD
    MOV CL,PW
    XOR CH,CH
    REPE CMPSB
    CMP CX,0
    JZ LD
    LC:
    LEA DX,ERR
    MOV AH,09H
    INT 21H
    JMP HK
    LD:
    LEA DX,MSG
    MOV AH,09H
    INT 21H
    mov ah,3eh
    int 21h ;关闭文件再打开，也可以用移动文件指针到文件开头
    LEA DX,FILE
    MOV AX,3D01H
    INT 21H
    MOV BX,AX
    CALL SET
    CALL SET1

    EXIT:
    MOV AH,4CH
    INT 21H
    MAIN ENDP

    SET PROC NEAR
    MOV CL,12H
    CLD
    LEA SI,PW
    ;MOV SI,13H 这句什么意思？错误？
    INC SI
    MOV DL,'*'
    I1:
    MOV AH,08H
    INT 21H
    CMP AL,0DH
    JZ R1
    MOV AH,02H
    CMP CL,0
    JNZ I2
    MOV DL,07H
    INT 21H
    JMP I1

    I2:
    MOV [SI],AL
    INT 21H
    INC SI
    DEC CL
    JMP I1
    R1:
    SUB PW,CL
    RET
    SET ENDP

    SET1 PROC NEAR
    LEA DX,PW
    MOV CL,PW
    AND CX,0FFH
    MOV AH,40H
    INT 21H
    mov ah,3eh
    int 21h ;关闭文件
    RET
    SET1 ENDP

    CODE ENDS
    还有错误
    温馨提示：答案为网友推荐，仅供参考
    其他回答
    第1个回答  2011-06-30
    #include<stdio.h>

    struct TData
    {
    int XueHao;
    char name[10];
    float L1,L2,L3,L;
    };

    void main()
    {
    int N=100; /*总共最多100名学生*/

    struct TData data[100];
    int sum=0; /*实际学生数目*/
    int sumbyl[11]; /*分段统计人数，按0-100分10段,100分单独1段*/
    int i,l;

    //输入学号、姓名、平时、期中和期末成绩，学号<0表示输入结束
    while(1)
    {
    printf("please intput XueHao:");

    scanf("%d",&data[sum].XueHao);
    if(data[sum].XueHao<0)
    break;
    printf("please input data:");
    scanf("%s %f %f %f",data[sum].name,&data[sum].L1,&data[sum].L2,&data[sum].L3);

    data[sum].L=data[sum].L1*0.2f+data[sum].L1*0.3f+data[sum].L1*0.5f; /*计算总成绩*/
    l=data[sum].L/10;
    sumbyl[l]++; /*分类统计*/

    sum++;
    if(sum>=N)
    break;
    }

    /*打印学生成绩*/
    printf("xuehao name L1 L2 L3 L\n");
    for(i=0;i<sum;i++)
    {
    printf("%d %s %f %f %f %f\n",data[i].XueHao,data[i].name,data[i].L1,data[i].L2,data[i].L3,data[i].L);
    }
    printf("\n");

    /*打印分数段人数及所占比例*/
    printf("范围 人数 比例\n");
    for(i=0;i<10;i++)
    {
    printf("%d-%d %d %f\n",i*10,i*10+9,sumbyl[i],sumbyl[i]*1.0f/sum);
    }
    printf("100-100 %d %f\n",sumbyl[11],sumbyl[11]*1.0f/sum);
    }
    第2个回答  2011-06-30
    #include<stdio.h>

    struct TData
    {
    int XueHao;
    char name[10];
    float L1,L2,L3,L;
    };

    void main()
    {
    int N=100; /*总共最多100名学生*/

    struct TData data[100];
    int sum=0; /*实际学生数目*/
    int sumbyl[11]; /*分段统计人数，按0-100分10段,100分单独1段*/
    int i,l;

    //输入学号、姓名、平时、期中和期末成绩，学号<0表示输入结束
    while(1)
    {
    printf("please intput XueHao:");

    scanf("%d",&data[sum].XueHao);
    if(data[sum].XueHao<0)
    break;
    printf("please input data:");
    scanf("%s %f %f %f",data[sum].name,&data[sum].L1,&data[sum].L2,&data[sum].L3);

    data[sum].L=data[sum].L1*0.2f+data[sum].L1*0.3f+data[sum].L1*0.5f; /*计算总成绩*/
    l=data[sum].L/10;
    sumbyl[l]++; /*分类统计*/

    sum++;
    if(sum>=N)
    break;
    }

    /*打印学生成绩*/
    printf("xuehao name L1 L2 L3 L\n");
    for(i=0;i<sum;i++)
    {
    printf("%d %s %f %f %f %f\n",data[i].XueHao,data[i].name,data[i].L1,data[i].L2,data[i].L3,data[i].L);
    }
    printf("\n");

    /*打印分数段人数及所占比例*/
    printf("范围 人数 比例\n");
    for(i=0;i<10;i++)
    {
    printf("%d-%d %d %f\n",i*10,i*10+9,sumbyl[i],sumbyl[i]*1.0f/sum);
    }
    printf("100-100 %d %f\n",sumbyl[11],sumbyl[11]*1.0f/sum);
    }
    第3个回答  2011-06-29
    学习标记，佩服楼主
    第4个回答  2011-06-29
    太长了，没时间看

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