Fortran Do循环的上限 [英] Upper bounds of Fortran Do loops

问题描述

在下面的代码中，DO循环的上界在循环中被修改为

In the following code, the upper bounds of the DO loops are modified within the loop as

    integer :: i1, i2, n1, n2
    n1 = 4 ; n2 = 1

    do i1 = 1, n1
    do i2 = 1, n2
        print *, "i1 =", i1, "i2 =", i2
        n1 = 2
        n2 = 2
    enddo
    enddo

其中gfortran4.8和ifort14.0给出以下结果：

for which gfortran4.8 and ifort14.0 give the following result:

i1 =           1 i2 =           1
i1 =           2 i2 =           1
i1 =           2 i2 =           2
i1 =           3 i2 =           1
i1 =           3 i2 =           2
i1 =           4 i2 =           1
i1 =           4 i2 =           2

这表示边界在每个DO循环进入时是固定的尽管在循环内n1被修改为2，i1的上限固定为4）。这个行为与C / C ++的行为相反，其中相应的代码

This indicates that the bounds are fixed upon entry of each DO loop (i.e., the upper bound of i1 is fixed to 4 despite that n1 is modified to 2 inside the loop). This behavior is in contrast to that of C/C++, where the corresponding code

int n1 = 4, n2 = 1;

for( int i1 = 1; i1 <= n1; i1++ )
for( int i2 = 1; i2 <= n2; i2++ )
{
    printf( "i1 = %d i2 = %d\n", i1, i2 );
    n1 = 2;
    n2 = 2;
}

给予

i1 = 1 i2 = 1
i1 = 1 i2 = 2
i1 = 2 i2 = 1
i1 = 2 i2 = 2

这是合理的，因为定义了for循环（即begin，end和increment条件）。所以我的问题是：是否可以假设DO循环的上述行为是一般的（=由Fortran标准定义）或编译器依赖（即不同的编译器从ifort或gfortran可能表现不同）？

which is reasonable because of the definition of for loops (i.e., begin, end, and increment conditions). So my question is: Is it okay to assume that the above behavior of DO loops is general (= defined by the Fortran standard) or compiler-dependent (i.e. a different compiler from ifort or gfortran might behave differently)?

推荐答案

Fortran标准规定 DO 循环的迭代次数在开始时是固定的使用公式（至少高达F95，当能力被移除以使用非整数循环控制表达式，如francescalus在注释中所指出的）。

The Fortran standards specify that the number of iterations of a DO loop is fixed on commencement of the loop, using the formula (at least up to F95, when ability was removed to use non-integer loop control expressions, as noted by francescalus in comments)

iterations = MAX(0,INT((final-value - initial-value + step-size)/step-size)

这样做的一个结果是你看到的行为：更改任何用于控制循环体内循环的值不会改变将要执行的迭代次数。

One consequence of this is the behaviour you are seeing: changing any of the values use to control the loop within the loop body does not change the number of iterations that will be executed.

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