Lina-金辉 2022-05-11 06:47 采纳率: 50%
浏览 108
已结题

编写C++代码实现二进制浮点数输入,显示其十进制数值结果,所有5类(可 以是0,非正规化数,正规化数,无穷,NaN)浮点结果都要进行测试和结果显示输入为一个32位的数据表示一个32位的二进制浮点数

编写C++代码实现二进制浮点数输入,显示其十进制数值结果,所有5类(可以是0,非正规化数,正规化数,无穷,NaN)浮点结果都要进行测试和结果显示。输入为一个32位的数据表示一个32位的二进制浮点数

  • 写回答

2条回答 默认 最新

  • 赵4老师 2022-05-11 09:20
    关注

    仅供参考:

    #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:
    //
    
    
    
    
    // https://docs.microsoft.com/en-us/cpp/c-runtime-library/format-specification-syntax-printf-and-wprintf-functions?view=vs-2019
    // https://docs.microsoft.com/zh-cn/cpp/porting/visual-cpp-change-history-2003-2015?view=vs-2019
    //
    // Starting in Visual Studio 2015, if the argument that corresponds to a floating-point conversion specifier (a, A, e, E, f, F, g, G) is infinite,
    // indefinite, or NaN, the formatted output conforms to the C99 standard. This table lists the formatted output:
    //   Value           Output
    //   --------------  ---------
    //   infinity        inf
    //   Quiet NaN       nan
    //   Signaling NaN   nan(snan)
    //   Indefinite NaN  nan(ind)
    // Any of these values may be prefixed by a sign. If a floating-point type conversion specifier character is a capital letter,
    // then the output is also formatted in capital letters. For example, if the format specifier is %F instead of %f, an infinity
    // is formatted as INF instead of inf. The scanf functions can also parse these strings, so these values can make a round trip
    // through printf and scanf functions.
    //
    // Before Visual Studio 2015, the CRT used a different, non-standard format for output of infinite, indefinite, and NaN values:
    //   Value                           Output
    //   ------------------------------  -------------------------
    //   + infinity                      1.#INF random-digits
    //   - infinity                      -1.#INF random-digits
    //   Indefinite (same as quiet NaN)  digit .#IND random-digits
    //   NaN                             digit .#NAN random-digits
    // Any of these may have been prefixed by a sign, and may have been formatted slightly differently depending on field width and precision,
    // sometimes with unusual effects. For example, printf("%.2f\n", INFINITY) would print 1.#J because the #INF would be "rounded" to 2 digits of precision.
    #include <stdio.h>
    #include <float.h>
    int main() {
        int r;
        double d1=-1.0,d2=-2.0;
        printf("input two double (d1 d2):");
        r=scanf("%lf%lf",&d1,&d2);
        printf("scanf return:%d d1:%lg d2:%lg\n",r,d1,d2);
        printf("_finite(d1):%d\n",_finite(d1));
        printf("_fpclass(d1):0x%04X\n",_fpclass(d1));
        return 0;
    }
    // input two double (d1 d2):nan 22
    // scanf return:2 d1:nan d2:22
    // _finite(d1):0
    // _fpclass(d1):0x0002
    //
    // input two double (d1 d2):nan(snan) 22
    // scanf return:2 d1:nan(snan) d2:22
    // _finite(d1):0
    // _fpclass(d1):0x0002
    //
    // input two double (d1 d2):nan(ind) 22
    // scanf return:2 d1:-nan(ind) d2:22
    // _finite(d1):0
    // _fpclass(d1):0x0002
    //
    // input two double (d1 d2):-inf 22
    // scanf return:2 d1:-inf d2:22
    // _finite(d1):0
    // _fpclass(d1):0x0004
    
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

问题事件

  • 系统已结题 5月21日
  • 已采纳回答 5月13日
  • 创建了问题 5月11日

悬赏问题

  • ¥50 关于#html5#的问题:H5页面用户手机返回的时候跳转到指定页面例如(语言-javascript)
  • ¥15 无法使用此凭据登录,因为你的域不可用,如何解决?(标签-Windows)
  • ¥15 yolov9的训练时间
  • ¥15 二叉树遍历没有报错但无法正常运行
  • ¥15 在linux系统下vscode运行robocup3d上场球员报错
  • ¥15 Python语言实验
  • ¥15 SAP HANA SQL 增加合计行
  • ¥20 用C#语言解决一个英文打字练习器,有偿
  • ¥15 srs-sip外部服务 webrtc支持H265格式
  • ¥15 在使用abaqus软件中,继承到assembly里的surfaces怎么使用python批量调动