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c++arraysnew-operator

Creating an array using new without declaring size

This has been bugging me for quite some time. I have a pointer. I declare an array of type int. 

int* data;
data = new int[5];

I believe this creates an array of int with size 5. So I'll be able to store values from data[0] to data[4].

Now I create an array the same way, but without size.

int* data;
data = new int;

I am still able to store values in data[2] or data[3]. But I created an array of size 1. How is this possible?

I understand that data is a pointer pointing to the first element of the array. Though I haven't allocated memory for the next elements, I still able to access them. How?

Thanks.

Solution

  • Normally, there is no need to allocate an array "manually" with new. It is just much more convenient and also much safer to use std::vector<int> instead. And leave the correct implementation of dynamic memory management to the authors of the standard library.

    std::vector<int> optionally provides element access with bounds checking, via the at() method.

    Example:

    #include <vector>
int main() {
    // create resizable array of integers and resize as desired
    std::vector<int> data; 
    data.resize(5);
    // element access without bounds checking
    data[3] = 10;
    // optionally: element access with bounds checking
    // attempts to access out-of-range elements trigger runtime exception
    data.at(10) = 0; 
}

    The default mode in C++ is usually to allow to shoot yourself in the foot with undefined behavior as you have seen in your case.

    For reference:

    Also, in the second case you don't allocate an array at all, but a single object. Note that you must use the matching delete operator too.

    int main() {
    // allocate and deallocate an array
    int *arr = new int[5];
    delete[] arr;
    // allocate and deallocate a single object
    int *p = new int;
    delete p;
}

    For reference: