浮动整数铸造？ [英] Float To Integer Casting?

问题描述

我知道浮点数不能完全代表每个数字，所以有些错误肯定会发生。

但最近我遇到了一个问题，我没有得到解释的权利。

请逐步解释转换如何影响输出。

小数位数是如何截去的错误的答案？

  #include< stdio.h> 
＃include< stdlib.h> 
＃include< math.h> 
 
 int main（）
 {
 int x; 
 int y = 0; 
 int n = 0; 
 
 printf（输入你想要的八进制数等于：）; 
 
 scanf（％d，& x）; （x> 0）
 {
 y = y +（x％8）* pow（10，n）; 
 
。 
 
 printf（％d\\\
，y）; 
 
 x = x / 8; 
 n = n + 1; 
} 
 
 printf（八进制等效为：％d，y）; 
返回0; 
 
 $ / code> 

当我输入查找八进制等值的 1701 。它给出答案 3244 。



我想知道的是程序如何工作？

程序如何打印 3244 而不是 3245 。



但答案是 3245 。



我检查了 x％8 是否正常工作。



当我将 y 数据类型更改为浮点数时，
为什么如此？

请逐步解释转换如何影响输出。

如何截断小数位给我错误的答案？

解决方案

当您从浮点类型转换为整数类型时，机器将截断小数位，而不是整数。

I know that floating point cannot represent every number exactly so some error is bound to happen.

But recently I have encountered a problem and I am not getting the explanation right.

Please explain me step by step how the conversion affected the output.

How did truncating decimal places give me the wrong answer?

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

int main()
{
   int x;
   int y = 0;
   int n = 0;

   printf("Enter The Number You Want To Octal Equivalent Of : ");

   scanf("%d",&x);

   while(x>0)
   {
      y = y + (x%8)*pow(10,n);

      printf("%d\n" , y);

      x = x/8 ;
      n = n + 1;
   }

   printf("Octal Equivalent is : %d" , y);
   return 0;
}

When I enter find octal equivalent of 1701. It gives answer 3244.

What I want to know is how does the program working?
How is the program printing 3244 and not 3245.

But the answer is 3245.

I have checked that x%8 is correctly working .

When I change the y data type to float, it works . Why so?

Please explain me step by step how the conversion affected the output.
How did truncating decimal places give me the wrong answer?

解决方案

When you convert from a floating-point type to an integer type, the machine will truncate the decimal places, not round.

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