Segmentation fault when using malloc to create array of pointers to pointers

Question

I'm having an issue with dynamically creating a "multi-dimensional array". I've read 6.14 on the comp.lang.c FAQ, and am following the code that was listed there.

            cache_array = malloc(cm_blks * sizeof(int *));
            if (cache_array = NULL) {
                    fprintf(stderr, "out of memory\n");
                    exit(1);
            }

            for (i = 0; i < cm_blks; i++) {
                    cache_array[i] = malloc(6 * sizeof(int));
                    if (cache_array[i] == NULL) {
                            fprintf(stderr, "out of memory\n");
                            exit(1);

                    }
            }

The variable cm_blks is an integer, in my test case equal to 8. cache_array is initialized as:

    int **cache_array;

The code compiles fine, but I get a segmentation fault at the second malloc line when I run the output.

hmjd · Accepted Answer · 2013-04-25 20:59:50Z

3

This is not an equality check but is an assignment:

if (cache_array = NULL)

which sets cache_array to NULL and does not enter the if branch as the result of the assignment is, essentially, false. The code then continues to dereference a NULL pointer.

Change to:

if (cache_array == NULL)

or:

if (!cache_array)

answered Apr 25, 2013 at 20:59

hmjd

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2 Comments

K Scott Piel Over a year ago

Good catch. It looks odd to many people, but it's often a good idea to place constants ahead of variables in conditional tests... for example: if( NULL = cache_array ) would have caught the error as an attempt to assign a variable to a constant. It's a handy way to avoid these kind of subtle errors.

user2321508 Over a year ago

Wow, I completely missed that! Thank you very much for the second pair of eyes!

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