从double到unsigned int的转换 [英] conversion from double to unsigned int

问题描述

我将Double值转换为int值时遇到问题。我尝试运行下面的代码：

$ pre $ $ $ $ $ $ $ $ $ $ $ $ -05;
cout<< （1 / val_d）< ENDL;
unsigned int val_ui =（unsigned int）（1 / val_d）;
cout<< val_ui<< ENDL;





$ p从double到int的转换可能会删除小数部分，但整数部分应保持为

我得到的输出是：
16000
15999

<那么为什么这里的o / p不同呢？
这只发生在fedora上。在Windows和Ubuntu上，它工作正常。 （两个输出都是16000）

我调整了上面的代码，得到了以下结果：

  int main（）
 {
 double val_d = 6.25e-05; 
 cout<< （1 / val_d）< ENDL; 
 double val_intermediate =（1 / val_d）; 
 cout<< val_intermediate<< ENDL; 
 unsigned int val_ui =（unsigned int）val_intermediate; 
 cout<< val_ui<< ENDL; 
 
 
 
 
 
 $ b新的输出是
 16000 
 16000 
 16000  
 
解决方案
当源文本6.25e-05被解释为十进制数字并转换为双精度，因为浮点值的精度有限，每个位的值都是2的幂，而不是十进制数，所以不完全可以表示。最接近6.25e-5的IEEE 754双精度值是6.25000000000000013010426069826053208089433610439300537109375e-05，或者在十六进制浮点型0x1.0624dd2f1a9fcp-14中。
 
 
当这个倒数倒数时，确切的数学结果又不能精确表示，所以必须再次舍入。最接近的双精度值是16000或0x1.f4p + 13。 
 
 
 C ++标准允许实现比标称类型更精确地计算浮点表达式。一些实现使用扩展精度，特别是Intel的80位浮点类型，它有一个64位有效数字。 （常规双精度有53位有效数）。在这个扩展精度下，倒数是0xf.9fffffffffffe89p + 10或15999.99999999999966693309261245303787291049957275390625。
 
 
显然，当扩展精度结果被截断为一个整数，结果是15999. 
 
 
将long-double结果舍入为double将产生16000.（可以通过显式强制转换来实现double ;您不需要将中间值赋给双重对象。）
 
I am facing a problem with the conversion of value from Double to int. I try to run the following code :
int main()
{
    double val_d= 6.25e-05;
    cout << (1/val_d) << endl;
    unsigned int val_ui = (unsigned int ) (1/val_d);
    cout << val_ui << endl;
}
conversion from double to int may remove decimal part but integer part should remain
as it is ?

The output i get is : 
16000
15999

so why is the o/p different here ?
This is happening only on fedora. On windows and Ubuntu it works fine. ( Both output are 16000)

I tweaked the above code and got the following results :
int main()
{
  double val_d= 6.25e-05;
  cout << (1/val_d) << endl;
  double val_intermediate =  (1/val_d) ;
  cout << val_intermediate << endl;
  unsigned int val_ui = (unsigned int ) val_intermediate;
  cout << val_ui << endl;

}
NEW OUTPUT is
16000
16000
16000
解决方案
When the source text "6.25e-05" is interpreted as a decimal numeral and converted to double, it is not exactly representable, because floating-point values have limited precision, and each bit has a value that is a power of two, not a decimal digit. The IEEE 754 double-precision value that is nearest to 6.25e-5 is 6.25000000000000013010426069826053208089433610439300537109375e-05, or, in hexadecimal floating-point, 0x1.0624dd2f1a9fcp-14.

When the reciprocal of this is taken, the exact mathematical result is again not exactly representable, so it must be rounded again. The nearest double-precision value is 16000 or 0x1.f4p+13.

The C++ standard allows implementations to evaluate floating-point expressions with more precision than the nominal type requires. Some implementations use extended precision, notably Intel's 80-bit floating-point type, which has a 64-bit significand. (Regular double precision has a 53-bit significand.) In this extended precision, the reciprocal is 0xf.9fffffffffffe89p+10 or 15999.99999999999966693309261245303787291049957275390625.

Obviously, when the extended-precision result is truncated to an integer, the result is 15999.

Rounding the long-double result to double would produce 16000. (You can do this with an explicit cast to double; you do not need to assign the intermediate value to a double object.)

                        
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