提高由整数商定义的变量的精度

问题描述

假设我有以下程序：

 程序derp 
隐式无
整数，参数： ：ikind = selected_real_kind（18）
 real（kind = ikind）:: a = 2.0 / 3.0 
 print *，a 
结束程序derp 
  

程序 derp 输出 0.6666666865348815917 ，这显然不是18位数的精度。但是，如果我使用相同的方法和定义 a = 2.0 和 b = 3.0 ，那么 define c = a / b 我得到的输出是 0.666666666666666666685 ，这很好。我怎样才能定义一个变量作为整数的商，并让它存储来自 selected_real_kind ？

解决方案

尝试： real（kind = ikind）:: a = 2.0_ikind / 3.0_ikind

原因在于，LHS的精度很高，代码示例中的RHS不是2.0 / 3.0。 Fortran以单精度进行计算，然后将结果分配给LHS。由于LHS的精度很高，因此RHS一侧的计算精度并不高。 digits_kind 是指定常量类型的方法 digits 。

Say I have the following program:

program derp
    implicit none
    integer, parameter :: ikind = selected_real_kind(18)
    real (kind = ikind) :: a = 2.0 / 3.0
    print*, a
end program derp

The program derp outputs 0.6666666865348815917, which is clearly not 18 digits of precision. However, if I define a=2.0 and b=3.0 using the same method and then define c=a/b I get an output of 0.666666666666666666685, which is good. How do I just define a variable as a quotient of integers and have it store all the digits of precision I want from selected_real_kind?

解决方案

Try: real (kind = ikind) :: a = 2.0_ikind / 3.0_ikind

The reason is while the LHS is high precision, the RHS in your code example, 2.0 / 3.0, is not. Fortran does that calculation in single precision and then assigns the result to the LHS. The RHS side isn't calculated in higher precision because the LHS is high precision. digits_kind is the way of specifying the type of a constant digits.

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