Questions on delete[] operator when pointer to an array is involved

Question

The following code compiles but runs with mistake:

int main() {
    int (*d)[2] = new int[3][2];
    // do something
    delete [] *d; // this is wrong
    delete [] *(d+1); // this is also wrong
    //delete [] d; // this works
    return 0;
}

I do not know why "delete [] *d" does not work because *d seems to be a pointer to a chunk containing 2 integers and delete[] should destroy that chunk.

Furthermore, I am not sure whether "delete [] d" is enough to release all six elements since two-dimension array is involved here.

Thanks!

delete what you new, and only what you new. The new-expression here returned d, not *d and not *(d+1). — aschepler
– aschepler, Commented Oct 10, 2016 at 19:47
int (*d)[2] says "*d is an array of two ints". An array is not a pointer. — molbdnilo
– molbdnilo, Commented Oct 10, 2016 at 20:09

AnT stands with Russia · Accepted Answer · 2016-10-10 20:02:06Z

4

The memory you allocated is allocated as one continuous block (or chunk) of memory that stores an array obejct of int[3][2] type. It should be freed as one block

delete [] d;

That's all you need to do.

Freeing just some partial sub-block of a whole block is not supported by C++ dynamic memory management mechanisms.

Your delete [] *d is indeed wrong - the behavior is undefined. However, it has a good chance of "working" in practice since int (*)[2] pointer d and [decayed] int * pointer *d point to the same spot in memory, and the underlying types are trivial. If it "works", it should typically have the same effect as delete [] d;, i.e. free the entire 2D array.

edited Oct 10, 2016 at 20:02

answered Oct 10, 2016 at 19:47

AnT stands with Russia

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Comments

Jean-François Fabre · Accepted Answer · 2016-10-10 19:47:06Z

0

int[3][2] is not a pointer of pointers but just a multi-dimensional array, allocated as a single pointer.

You cannot delete "lines" of it because there are no lines. Memory is contiguous.

answered Oct 10, 2016 at 19:47

Jean-François Fabre♦

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