十进制小数转十六进制
请问这两道
怎么做呀？为什么十进制-4.5是C803而不是C800？
不知道你们课堂上是怎么讨论的，建议你去翻翻课堂笔记。
十进制-4.5是C803而不是C800，我猜二进制的结构大约是符号位占1位，整数位占3位的意思。据此，可以大致推断：
-13.25 对应的二进制是 1110 1010 0000 0100，即EC04
-0.1875 对应的二进制是 1000 0011 0000 0011，即8303
0.375 对应的二进制是 0000 0110 0000 0011，即0603
2020 对应的二进制是 0111 1110 0100 1011，即774B
如果这个推断正确的话，十六进制7802 等价于十进制3.75，倒是正好符合推论。E40A表示-800，4305表示16.75，也可以说得通。但是6BCD和7A0D，最后都是13，就有点无从下手了。
以上纯属瞎猜，仅供参考。总之，解决问题还得要依赖课堂上的讨论或者课本上的东西。
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:
//
是不是整数数部分+小数部分的，主要是你的课堂上讲的规则是什么？
011.1 1000 0000 0010
3.75
111.0 0100 0000 1010
7.125
010.0 0011 0000 0101
2.09375
011.0 1011 1100 1101
3+1/4+1/16+1/32+1/64+1/128=3.3671875
-4 .5
1100 1000 0000 0011