如何分离整数部分和小数部分？ [英] how seperate integer part and fractional part?

问题描述

我想分开实数的整数部分和小数部分。

即使输入了56.345 int部分将显示为56，frctional部分显示为345.以下是我尝试的代码：

Hi,

I want to seperate the integral part and fractional part of a real number.
That is if 56.345 is entered the int part will be shown as 56 and frctional part as 345. Here is the code I tried:

#include<iostream.h>
#include<conio.h>
int main()
{
float x;
int a[15],b[15];
clrscr();
cout<<"\nEnter the number:";
cin>>x;
int y=x;
float z=(x-y);
cout<<"\nInteger part    :"<<y;
cout<<"\nFractional part :"<<z;
getch();
return 0;
}

输出结果如下......

The output is as follows...

case 1:
Enter the number:24.41

Integer part    :24
Fractional part :0.41

case 2:
Enter the number:56.345

Integer part    :56
Fractional part :0.345001

在第二种情况下，我无法获得正确的答案。我可以通过使用char数组解决这个问题，但我不想使用那种方法。



（编辑使文本看起来像英文而不是阿里-G湿梦想）。

In the second case I can't obtain the correct answer. I can solve this by using a char array but I dont want to use that method.

(Edited to make the text look like English and not an Ali-G wet dream).

推荐答案

您的问题是计算机不能完全代表56.345并最终将其存储为56.3450001。这是一个不幸的事实，它需要无限多的位来存储二进制的任意十进制数。



（实际上它需要无穷多的十进制数字作为好吧，但这是另一个故事。）



最快的方法是将数字转换为固定点，你可以存储整数：

Your problem is that the computer can't exactly represent 56.345 and ends up storing it as 56.3450001. It's an unfortunate fact of life that it takes an infinite number of bits to store a arbitrary decimal number in binary.

(Actually it takes an infinite number of decimal digits as well, but that's another story).

The quickest way around this is to convert the number to fix point, which you can store in an integer:

double x = 0.0;
std::cin >> x;
int y  = int( x * 100000 );

std::cout << "Integral part:   " << y / 100000 << std::endl
          << "Fractional part: " << y % 100000 << "/10000" << std::endl;

这不适用于不适合整数的大数字。例如，乘以10,000会使整数中大约一半的位用于保持小数部分，从而使您的范围更大。最终你必须弄清楚你可以接受的范围和不准确度。并且可以使用它。



干杯，



Ash

This won't work with big numbers that won't fit in an integer. Multiplying by 10,000 for example makes about half the bits in the integer be used for holding the fractional part so you loose even more range. Ultimately you've got to work out what range and inaccuracy are acceptable to you and go with it.

Cheers,

Ash

要将整数部分转换为 int ，小数部分为数字 - 整数部分，围绕它使用iostream操纵器（或C printf（）函数）：

To get the integer part convert to int, the fractional part is number - integer part, to round it use the iostream manipulators (or C printf() functions):

#include <iostream>
#include <iomanip>
int main()
{
    float f=1;
    while (f)
    {
        std::cin >> f;
        int i = (int)f;
        std::cout << i << '\t' << std::setprecision(3) << f - i << std::endl;
    }
    return 0;
}

欢呼，

AR

cheers,
AR

将其转换为文本，以十进制分割并转换回数字。



非常简单。



您需要做的就是在X有效数字后删除后端。

Convert it to text, split at the decimal and convert back to numbers.

Quite simple really.

All you need to do is remove the back end after X significant figures.

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