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Deleting a dynamically allocated 2D array [duplicate]

So I'm used to memory management in C where free(pointer) will free up all space pointed to by pointer. Now I'm confusing myself when attempting to do something simple in C++.

If I have a 2D array of doubles allocated in a manner similar to this

double** atoms = new double*[1000];
for(int i = 0; i < 1000; i++)
  atoms[i] = new double[4];

what would be the correct method of freeing the memory on the heap allocated by new?

My thoughts were originally this (because my brain was thinking in C):

for(int i = 0; i < 1000; i++)
  delete atoms[i];
delete atoms;

But I had forgotten the existence of the delete[] operator so I believe the correct method is as follows:

for(int i = 0; i < 1000; i++)
  delete[] atoms[i];
delete[] atoms;

Is it important to understand the difference between the delete and delete[] operators? Or can I just assume that whenever I allocate an array with ptr = new x[] I must also deallocate it with delete[] ptr?

like image 767
Alex Avatar asked Jun 08 '15 23:06

Alex


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1 Answers

In reality, an array of pointers pointed to by a pointer is still an array of integral data types or numbers to hold the memory addresses. You should use delete[] for both.

Also, yes, a new[] implies a delete[].

When you create an array of arrays, you're actually creating an array of numbers that happen to hold the memory address for another array of numbers. Regardless, they're both arrays of numbers, so delete both with delete[].

http://coliru.stacked-crooked.com/a/8a625b672b66f6ce

#include <iostream>

int main() {

    //Hey, pointers have a finite size, no matter the indirection level!
    std::cout << "sizeof(int*): " << sizeof(int*) << std::endl;
    std::cout << "sizeof(int**): " << sizeof(int**) << std::endl;
    std::cout << "sizeof(int***): " << sizeof(int***) << std::endl;

    //Create an array of pointers that points to more arrays
    int** matrix = new int*[5];
    for (int i = 0; i < 5; ++i) {
        matrix[i] = new int[5];
        for (int j = 0; j < 5; ++j) {
            matrix[i][j] = i*5 + j;
        }
    }

    //Print out the matrix to verify we have created the matrix
    for (int j = 0; j < 5; ++j) {
        for (int i = 0; i < 5; ++i) {
            std::cout << matrix[j][i] << std::endl;
        }
    }

    //Free each sub-array
    for(int i = 0; i < 5; ++i) {
        delete[] matrix[i];   
    }
    //Free the array of pointers
    delete[] matrix;

    return 0;
}
like image 155
CinchBlue Avatar answered Oct 02 '22 18:10

CinchBlue