Why does this code not result in a memory leak? [duplicate]

Question

I checked the following code in C++ with valgrind with --leak-check=full and it says no memory leak. Why is that?

char *p = new char[256]; delete p; 

new[] should be matched by delete[] as far as I know.

Answer 1

Although it's undefined behaviour as @KillianDS says, the difference probably relates to the fact that both delete and delete[] free the underlying memory. The point of delete[] is that the destructors for each object in the array are called before the memory is freed. Since char is a POD and doesn't have a destructor, there isn't any effective difference between the two in this case.

You definitely shouldn't rely on it, however.

Answer 2

delete and delete[] will be equal only if the p points to basic data types, such as char or int.

If p points to an object array, the result will be different. Try the code below:

class T { public:     T() { cout << "constructor" << endl; }     ~T() { cout << "destructor" << endl; } };  int main() {     const int NUM = 3;     T* p1 = new T[NUM];      cout << p1 << endl;     //  delete[] p1;     delete p1;      T* p2 = new T[NUM];     cout << p2 << endl;     delete[] p2; } 

By using delete[] all the destructors of T in the array will be invoked. By using delete only p[0]'s destructor will be invoked.