在C中释放二维数组

Question

Please bear with me as this is probably a very simple question but I am very new to C. I am trying to malloc a specific array and then free it. However, the line:

    M = malloc(N*sizeof(double *));

...doesn't work. Can somebody please explain to me why this is not working and what the solution would be? Many thanks in advance.

Answer 1

You should free all positions, not just the first. See this example I have wrote.

As mentioned from @Grijesh, the allocation is also wrong. The example covers allocation too. Moreover, I suggest you not to cast the return of malloc ( more ).

You have to think the 2D array, as a 1D array, where every cell of it is a pointer to a 1D array. A picture might help: http://gsamaras.files.wordpress.com/2014/04/array2d-n.png Here, 1D array that holds the pointers is to the left and every cell of it, points to another 1D array. How many 1D arrays to the left? As many cells as you have in the left array.

Btw, Nelly I think this is not a silly question, it's something that gets beginners into trouble. ;)

EDIT: About your new code, you had to have the same definition and declaration for matrix_free, as well as, call it as you should. What's definition, etc. ?? Answer .

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

#define Nmax 9000

double **makeMatrix(int N);
void free_matrix(double **M,int N);

int main(void) {
    int N;

    for (N = 2; N < Nmax; N *= 2) {
        double **L = makeMatrix(N);
        printf("yes \n");
        free_matrix(L, N);
        printf("woo \n");
    }
    return 0;
}

double **makeMatrix(int N) {
    int i, j;
    double **M;

    M = malloc(N * sizeof(double *));
    for (i = 0; i < N; i++)
        M[i] = malloc(N * sizeof(double));

    for (i = 1; i < N; i++) {
        for (j = 1; j < N; j++) {
            M[i][j] = (i) * (j) * M_PI / N;
        }
    }

    return (M);
}

void free_matrix(double **M, int N) {
    int i;
    for (i = 1; i <= N; i++) {
        free(M[i]);
    }
    free(M);
}

And then I receive the youwho output. :) But, it will stop at a certain point, because NMAX is too big! Not only NMAX is too big, but N grows really fast ( N *= ). Have you done the math in a piece of paper? Too big numbers. For example, if I do N += , then, I can go until NMAX = 9000 . Debug tip : How do I know in which loop it reaches? I printed out the counter of the loop, like this

printf("woo %d\n",N);

Of course, if you feel sure for yourself, then I suggest you learning the debugger.