Is delete and delete[] equivalent for basic data types
c++, delete-operator
Solution
If you use `new` you need to use `delete`
If you use `new []` you have to use `delete []`
They mean different things. `double* d = new double();` means allocate and construct a double type. `delete d;` means unallocated and destruct a single double type. `double* d = new double[NUM_VALUES];` means allocate `NUM_VALUES` doubles and `delete [] d` means unallocated each of of the `NUM_VALUES` doubles allocated.
Problem
So during a code review, a colleague of mine used a `double* d = new double[foo];` and then called `delete d`. I told them that they should change it to `delete [] d`. And they stated that the compiler doesn't need that for basic data types. I disagreed. So I thought I'd prove my point with an experiment: ``` #define NUM_VALUES 10000 int main(int argc,char** argv) { int i = 0; while (true) { std::cout << i << "|"; double* d = new double[NUM_VALUES]; std::cout << ((void*)d) << std::endl; for (int j = 0; j < NUM_VALUES; j++) d[j] = j; delete d; i++; } return 0; } ``` Not only does the memory usage not grow, but d is allocated to the same place every time! (Visual Studio 2010). Is this a quirk of the visual studio compiler? Or is this part of the standard?