Question

我正在尝试将在C中初始化的4D数组传递给FORTRAN函数。根据前一个question的建议，我首先使用单个malloc将数据数组分配为连续数据，然后分配指针指针，使它们最终指向数据数组。此方法适用于3D阵列。但是，当我转到4D阵列时，我没有得到FORTRAN中预期的值。当我为C数组malloc内存时，是否需要改变某些事情＆gt; 3个维度或者alloc_4D_double函数中是否有错误？

代码摘录：

#include <stdio.h>
#include <stdlib.h>

double f_function(double ****);

double ****alloc_4D_double(int wlen, int xlen, int ylen, int zlen, double *ary)
{
    int i,j,k;

    double ****multiDary = (double****)malloc(wlen*sizeof(double***));

    for (i = 0; i < wlen; i++) 
    {
        multiDary[i] = (double***)malloc(xlen*sizeof(double**));

        for (j = 0; j < xlen; j++) 
        {
            multiDary[i][j] = (double**)malloc(ylen*sizeof(double*));

            for (k = 0; k < ylen; k++) 
            {
                multiDary[i][j][k] = (double*)malloc(zlen*sizeof(double));
            }
        }
    }

    for (i = 0; i < wlen; i++) 
    {                     
        for (j = 0; j < xlen; j++) 
        {
            for (k = 0; k < ylen; k++) 
            {
                multiDary[i][j][k] = &ary[zlen*(k + ylen*(j + xlen*i))];            
            }
        }
    }

    return multiDary;
}

int main ( void ) {

    // create contiguous memory
    double *data1D = (double*)malloc(2 * 4 * 10 * 10 * sizeof(double));

    // map memory locations to allow use of array syntax
    double ****exposure4D = alloc_4D_double(2, 4, 10, 10, exposure1D);

    // assign a value
    data[0][1][4][4] = 0.777;

    // pass the pointer to FORTRAN
    f_funtion(data1D)

    return 0;
}

和

integer(4) function f_function(data)  BIND(C)
    use, intrinsic :: iso_c_binding    

    implicit none        
!
    real(C_DOUBLE)       , intent(in),      dimension(2,4,10,10) :: data

    // this value does not evaluate to 0.7777
    write(*,*), data(1,2,5,5) 

    // this value does evaluate to 0.7777
    write(*,*), data(1,1,9,2) 

end function f_function

将4D C数组传递给FORTRAN函数

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