在C ++中将double转换为int而不舍入错误 [英] Convert double to int in C++ without round down errors

问题描述

我有以下代码将 double 转换为 int ：

  double dblValue = 7.30; 
 int intValue =（int）（dblValue * 100）; //我想intValue存储极端730; 
 cout<< intValue; 
    729   
 我知道编译器读取dblValue为7.2999999，然后将它转换为int。 
  
 我的问题是：可以通过防止舍入错误？
 
 
 
如果你的解决方案避免使用C ++ 11或其他预定义的函数，这将是最好的。我使用的唯一的预处理器指令是< iostream> 。
解决方案
在将不是整数（数学意义上的）的数字转换为整数时，不能防止舍入错误，唯一可以做的是尝试实现正确的舍入。
 
 
 
实现一个明智的（虽然不是完美的）四舍五入的最简单的方法如下：
  int intValue =（int）（dblValue <0≤dblValue-0.5：dblValue + 0.5）; 
  
当然，因为你的问题被标记为 c ++ 和铸造我无法阻止用c ++风格的cast替换你的c风格的cast：
  int intValue = static_cast< int>（dblValue< 0？dblValue-0.5：dblValue + 0.5）; 
  
 
I have the following codes to cast a double into an int:
double dblValue = 7.30;
int intValue = (int)(dblValue*100);  //I want intValue to store extactly 730;
cout << intValue;
Output: 729
I know that the compiler is reading dblValue as 7.2999999 before casting it to int.
My question is: Is it possible to cast it as 730 by preventing the round down errors?

It would be best if your solution avoid using C++11 or other predefined functions. The only pre-processor directive I am using here is <iostream>.
解决方案
You cannot prevent rounding errors when converting a number that is not an integer (in the mathematical sense) to an integer, the only thing you can do is try to achieve proper rounding.

The easiest way to achieve a sensible (although not perfect) rounding this is the following:
int intValue = (int)(dblValue < 0 ? dblValue - 0.5 : dblValue + 0.5);
And of course, since your question is tagged both c++ and casting I cannot resist replacing your c-style cast with a c++ style cast:
int intValue = static_cast<int>(dblValue < 0 ? dblValue - 0.5 : dblValue + 0.5);


                        
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