What does 'delete [] _v;' mean?

Question

template<>
class CalcHashClass<const char*> {
public:
    CalcHashClass(const char* v) {
        _v = new char[strlen(v) + 1];
        strcpy(_v,v);
    }

    ~CalcHashClass() {
        delete [] _v;
    }

    int hashCode() {
        printf("This is 'template <> class CalcHashClass<const char*>'.\n");
        int len = strlen(_v);
        int code = 0;
        for (int i = 0; i < len; ++i)
            code += (int)_v[i];
        return code;
    }

private:
    char* _v;
};

For above code, I am not sure what delete [] _v; mean? In my understanding, _v is a pointer. And to delete it should use delet _v, am I right? What does the [] mean?

It means `_v` is an array. You can see the `[` and `]` in the `new` statement as well. — Drew Dormann, Mar 20 '13 at 14:40
I'm not sure how you even got to template specialization if you have not learned about raw-arrays and member variables. — dchhetri, Mar 20 '13 at 14:43

score 4 · Accepted Answer · answered Mar 20 '13 at 14:41

4

delete x will only delete the variable x whereas the [ ] indicate to the memory manager that an array was allocated. delete[] x will delete the allocated array.

answered Mar 20 '13 at 14:41

Suvarna Pattayil

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I like this answer, simple and easy to understand. Thanks! – Tom Xue Mar 20 '13 at 14:47

score 3 · Answer 2 · answered Mar 20 '13 at 14:41

Look at how the object that _v points to was allocated:

_v = new char[strlen(v) + 1];

It uses the new char[...] syntax, meaning that it is dynamically allocating an array of char. To destroy a dynamically allocated array, you need to use delete[].

The reason that this is necessary is because the compiler cannot tell that _v is pointing at an array from the type of _v alone. Since _v is just a char*, there's no reason you couldn't point it at a single dynamically allocate char. So delete[] tells the compiler that it needs to find out exactly how much space was allocated here and then deallocate all of it.

score 1 · Answer 3 · edited May 23 '17 at 12:23

delete[] is called the "array delete operator". It deletes an array allocated using new[].

It is important to note that it is not allowed to use array delete with non-array new and vice versa.

For more information, see Why do we even need the "delete[]" operator?

As a side note, your class violates the Rule of Three. You should provide (or disable) the copy constructor and copy assignment operator.

score 1 · Answer 4 · answered Mar 20 '13 at 14:40

When you have an array allocated using new [] you need to use delete [] if you use just delete with memory allocated using new [] you will have undefined behavior. In the code posted you are allocating using new [] in your constructor here:

_v = new char[strlen(v) + 1];

hence the call to delete [] later on in the destructor.

What does 'delete [] _v;' mean?

4 Answers4