随机数生成器和4字节vs. 8字节整数的溢出

问题描述

著名的线性同余随机数生成器，也称为最小标准使用公式

The famous linear congruential random number generator also known as minimal standard use formula

x(i+1)=16807*x(i) mod (2^31-1)

我想使用Fortran来实现.

I want to implement this using Fortran.

但是，正如数字食谱"所指出的那样，使用默认的Integer类型(32位)直接实现该公式将导致 16807 * x(i)溢出.

However, as pointed out by "Numerical Recipes", directly implement the formula with default Integer type (32bit) will cause 16807*x(i) to overflow.

因此，本书推荐Schrage的算法基于m的近似因式分解.仍然可以使用默认整数类型来实现此方法.

So the book recommend Schrage’s algorithm is based on an approximate factorization of m. This method can still implemented with default integer type.

但是，我想知道fortran实际上具有 Integer(8)类型，其范围是 -9,223,372,036,854,775,808 到 9,223,372,036,854,775,807 ，它比 16807 * x(i)是.

However, I am wondering fortran actually has Integer(8) type whose range is -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 which is much bigger than 16807*x(i) could be.

但书甚至说了以下一句话

but the book even said the following sentence

那么为什么我们不能只使用 Integer(8)类型直接实现公式?

So why can't we just use Integer(8) type to implement the formula directly?

推荐答案

是否可以使用8字节整数取决于您的编译器和系统.您可以使用 selected_int_kind 方法来获得一种具有一定范围的int.该代码可在我的64位计算机上编译并正常运行:

Whether or not you can have 8-byte integers depends on your compiler and your system. You can use the selected_int_kindmethod to get a kind of int that has a certain range. This code compiles on my 64 bit computer and works fine:

program ran
    implicit none
    integer, parameter :: i8 = selected_int_kind(R=18)
    integer(kind=i8) :: x
    integer :: i
    x = 100
    do i = 1, 100
        x = my_rand(x)
        write(*, *) x
    end do

    contains
        function my_rand(x)
            implicit none
            integer(kind=i8), intent(in) :: x
            integer(kind=i8) :: my_rand
            my_rand = mod(16807_i8 * x, 2_i8**31 - 1)
        end function my_rand
end program ran

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