将表示2D数组的指针传递给C ++中的函数

Question

http://www.neilstuff.com/guide_to_cpp/notes/Multi%20Dimension%20Arrays%20and%20Pointer%20Pointers.htm

According to this site, I should be able to use the following code:

double stuff[3][3];
double **p_stuff;
p_stuff = stuff;

But I get a complaint that the conversion is not allowed by assignment.

Am I doing something wrong?

I have an extern "C" type function that I want to pass this double stuff[3][3] to. So I think i need to make it a pointer, right?

Answer 1

Regarding the edit: to pass this double stuff[3][3] to a C function, you could

1) pass a pointer to the whole 2D array:

void dostuff(double (*a)[3][3])
{
// access them as (*a)[0][0] .. (*a)[2][2]
}
int main()
{
    double stuff[3][3];
    double (*p_stuff)[3][3] = &stuff;
    dostuff(p_stuff);
}

2) pass a pointer to the first 1D array (first row) and the number of rows

void dostuff(double a[][3], int rows)
{
// access them as a[0][0] .. a[2][2]
}
int main()
{
    double stuff[3][3];
    double (*p_stuff)[3] = stuff;
    dostuff(p_stuff, 3);
}

3) pass a pointer to the first value in the first row and the number of both columns and rows

void dostuff(double a[], int rows, int cols)
{
// access them as a[0] .. a[8];
}
int main()
{
    double stuff[3][3];
    double *p_stuff = stuff[0];
    dostuff(p_stuff, 3, 3);
}

(that this last option is not strictly standards-compliant since it advances a pointer to an element of a 1D array (the first row) past the end of that array)

If that wasn't a C function, there'd be a few more options!

Answer 2

Your assigned is flawed. p_stuff; is pointer to pointer to double whereas stuff is two dimensional array( array of arrays)

A single dimension array decays to the pointer to its first element. A 2 dimensional array decays to pointer to a single dimension array.

Try this

double stuff[3][3];
double (*p_stuff)[3]; // pointer to array of 3 int
p_stuff = stuff;

Answer 3

double ** p_stuff; corresponds to an array of pointer to double . double stuff[3][3] doesn't have any pointers - it's a 2D array of double.

Answer 4

Below is maybe your answer:

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

typedef unsigned int uint;
uint m[10][20];
uint **ppm;

int main() {
    int i;

    ppm = (uint **)m;
    for (i =0; i<10; ++i)ppm[i] = (uint *)&m[i];

    m[1][1] = 10;

    printf("0x%x vs 0x%x: %d vs %d\n", ppm,m, m[1][1], *(*(ppm+1)+1));

    return 0;
}

The result's on console screen:

0x6010a0 vs 0x6010a0: 10 vs 10

Answer 5

In addition to Cubbi's detailed answer, the following way is just natural to pass a 2-dim array to a function:

void dostuff(void *a1, int rows, int cols)
{
    double (*a)[cols] = (double (*)[cols]) a1;
    // access them as a[0][0] .. a[rows-1][cols-1];
}
int main()
{
    double stuff[3][3];
    dostuff(stuff, 3, 3);
}

You don't have to cast them in advance since the newest C++ (C++14) supports run-time sized arrays.