#include "ieee754.h"
#include <stdio.h>
#include <math.h>
//This program convert a floating number to its binary representation (IEEE754) in computer memory
int main() {
long double f, binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;
long int integer, binaryInt = 0;
long int p = 0, rem, temp;
printf("\nEnter floating number: ");
scanf("%Lf", &f);
//separate the integer part from the input floating number
integer = (int)f;
//separate the fractional part from the input floating number
frac = f - integer;
//loop to convert integer part to binary
while (integer != 0) {
rem = integer % 2;
binaryInt = binaryInt + rem *pow(10, p);
integer = integer / 2;
p++;
}
//loop to convert fractional part to binary
while (frac != 0) {
frac = frac * 2;
temp = frac;
binaryFrac = binaryFrac + fracFractor * temp;
if (temp == 1)
frac = frac - temp;
fracFractor = fracFractor / 10;
}
binaryTotal = binaryInt + binaryFrac;
printf("binary equivalent = %Lf\n", binaryTotal);
}
我正在尝试将浮点数转换为二进制表示（64位）。这段代码有效，但并不完美。例如，当我转换.575它给我0.100100但是当我使用这个网站http://www.exploringbinary.com/floating-point-converter/进行转换时，正确的输出应该是0.1001001100110011001100110011001100110011001100110011
我无法理解是什么让我的代码截断数字。任何人都可以帮我解决这个问题吗？我感谢您的帮助。
很多问题：
(int)提取long double的整数部分严重限制了范围。使用modfl(long double value, long double *iptr);
long double f;
long int integer;
//separate the integer part from the input floating number
// Weak code
integer = (int)f;
long double ipart;
long double fpart = modfl(f, &ipart);
long p; pow(10,p); - >一旦pow()超过某个值，p返回值的精度损失，（例25）。使用pow()和使用long double的功能也很奇怪。我期待powl()。fracFractor/10，long的精度有限。代码很奇怪，因为它试图将FP编号（可能是某种二进制格式）转换为二进制表示。它不应该在代码中的任何地方需要10。
建议简单的东西
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<float.h>
static void print_ipart(long double x) {
int digit = (int) (modfl(x/2, &x)*2.0) + '0';
if (x) {
print_ipart(x);
}
putchar(digit);
}
void print_bin(long double x) {
// Some TBD code
// Handle NAN with isnan()
// Handle infinity with isfinite()
putchar(signbit(x) ? '-' : '+');
long double ipart;
long double fpart = modfl(fabsl(x), &ipart);
print_ipart(ipart);
putchar('.');
while (fpart) {
long double ipart;
fpart = modfl(fpart * 2, &ipart);
putchar((int)ipart + '0');
}
putchar('\n');
}
int main() {
print_bin(-4.25);
print_bin(.575);
print_bin(DBL_MAX);
print_bin(DBL_MIN);
print_bin(DBL_TRUE_MIN);
}
产量
-100.01
+0.1001001100110011001100110011001100110011001100110011001100110011
+1111111111111111111111111111111111111111111111111111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.
+0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
+0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
这就是为什么这不太可行的原因：
fracFractor = 0.1
...
fracFractor = fracFractor/10
0.1不能以任何二进制浮点格式准确表示。你不能将0.1表示为2的负幂的倍数。将它除以10将使每一步都收集舍入误差。可能是你实际退出这个循环，因为你最终将重复分数与另一个重复分数进行比较。
这将严重限制您可以实现的目标：
binaryTotal = binaryInt + binaryFrac;
以浮点执行此操作将具有严重的限制 - 至少表示如上所述无法表示0.1。这就是为什么你得到的答案显示为二进制和十进制数字的混合。
要解决这个问题，您应该查看数字的各个位。为了保持解决方案的整体概念不变，最简单的方法是从你的分数中减去2的负功率（0.5,0.25等），测试它是否仍为正，并根据它构建一个字符串。然后对整数部分使用类似的逻辑。