Zeng-fh 2022-03-08 16:01 采纳率: 89.8%
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已结题

关于数据类型、数据输出的疑问

这段代码的第一个输出为什么后面两位出现了设定值中没有的值啊


#include <stdio.h>
main()
{
    float x=1234.567;
    double y=1234.5678;
    printf("x=%f,y=%f\n",x,y);
    printf("x=%6.3f,y=%10.3f\n",x,y);
}

img

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1条回答 默认 最新

  • 赵4老师 2022-03-08 16:40
    关注
    
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    int main() {
        float f;
        double d;
        char bs[65];
        char b[65];
        char s[80];
        unsigned char *p;
        char e[12];
        char *t;
        int ex;
        int flag;
    
        flag=0;
        while (1) {
            printf("Input a float point number or 0xXXXXXXXX or 0xXXXXXXXXXXXXXXXX:");fflush(stdout);
            rewind(stdin);
            fgets(s,80,stdin);
            if ('\n'==s[0]) return 1;
            if (1==sscanf(s,"0x%16I64X",(__int64 *)&d) && strlen(s)>11) {flag=2;break;}
            if (1==sscanf(s,"0x%8X"    ,(  int   *)&f))                 {flag=1;break;}
            if (1==sscanf(s,"%f"       ,           &f)
             && 1==sscanf(s,"%lf"      ,           &d))                 {flag=3;break;}
        }
        if (flag&1) {
            printf("f=%g\n",f);
            p=(unsigned char *)&f;
            printf("hex=%02X %02X %02X %02X\n",p[3],p[2],p[1],p[0]);
            ltoa(*(long *)&f,b,2);
            sprintf(bs,"%032s",b);
            printf("bin=%s\n",bs);
            printf("bin=%.1s %.8s   %s\n",bs,bs+1,bs+9);
            strncpy(e,bs+1,8);e[8]=0;
            ex=strtol(e,&t,2);
            printf("    %c %-4d-127 1.%s\n",(bs[0]=='0')?'+':'-',ex,bs+9);
            ex-=127;
            printf("    %c %-8d 1.%s\n",(bs[0]=='0')?'+':'-',ex,bs+9);
        }
        if (flag&2) {
            printf("\nd=%lg\n",d);
            p=(unsigned char *)&d;
            printf("hex=%02X %02X %02X %02X %02X %02X %02X %02X\n",p[7],p[6],p[5],p[4],p[3],p[2],p[1],p[0]);
            _i64toa(*(__int64 *)&d,b,2);
            sprintf(bs,"%064s",b);
            printf("bin=%s\n",bs);
            printf("bin=%.1s %.11s   %s\n",bs,bs+1,bs+12);
            strncpy(e,bs+1,11);e[11]=0;
            ex=strtol(e,&t,2);
            printf("    %c %-6d-1023 1.%s\n",(bs[0]=='0')?'+':'-',ex,bs+12);
            ex-=1023;
            printf("    %c %-11d 1.%s\n",(bs[0]=='0')?'+':'-',ex,bs+12);
        }
    
        return 0;
    }
    //Input a float point number or 0xXXXXXXXX or 0xXXXXXXXXXXXXXXXX:0x3FC0000000000000
    //
    //d=0.125
    //hex=3F C0 00 00 00 00 00 00
    //bin=0011111111000000000000000000000000000000000000000000000000000000
    //bin=0 01111111100   0000000000000000000000000000000000000000000000000000
    //    + 1020  -1023 1.0000000000000000000000000000000000000000000000000000
    //    + -3          1.0000000000000000000000000000000000000000000000000000
    //
    //Input a float point number or 0xXXXXXXXX or 0xXXXXXXXXXXXXXXXX:0x3E000000
    //
    //f=0.125
    //hex=3E 00 00 00
    //bin=00111110000000000000000000000000
    //bin=0 01111100   00000000000000000000000
    //    + 124 -127 1.00000000000000000000000
    //    + -3       1.00000000000000000000000
    //
    //Input a float point number or 0xXXXXXXXX or 0xXXXXXXXXXXXXXXXX:0.125
    //f=0.125
    //hex=3E 00 00 00
    //bin=00111110000000000000000000000000
    //bin=0 01111100   00000000000000000000000
    //    + 124 -127 1.00000000000000000000000
    //    + -3       1.00000000000000000000000
    //
    //d=0.125
    //hex=3F C0 00 00 00 00 00 00
    //bin=0011111111000000000000000000000000000000000000000000000000000000
    //bin=0 01111111100   0000000000000000000000000000000000000000000000000000
    //    + 1020  -1023 1.0000000000000000000000000000000000000000000000000000
    //    + -3          1.0000000000000000000000000000000000000000000000000000
    //
    //Input a float point number or 0xXXXXXXXX or 0xXXXXXXXXXXXXXXXX:
    //
    
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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  • 系统已结题 4月3日
  • 已采纳回答 3月26日
  • 创建了问题 3月8日

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