c语言整数转单浮点数输出

Question

uint最大整数根据I3E转成单精度浮点后16进制是0x4F7FFFFF,再转回整数是4294967040
为什么输出反而变大了?

#include <stdio.h>

int main(void) {
    float f1 = 0xffffffff;
    unsigned int umax = 0xffffffff;
    printf(" u is:%u\n",umax); // u is:4294967295
    printf("f1 is:%f\n",f1); //f1 is:4294967296.000000
    return 0;
}

Answer 1

float最长有效十进制位数=6
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:
//