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Does delete ptr differ from operator delete(ptr) only in this, that delete calls ptr destructor? Or in other words, does delete ptr first call a destructor of ptr and then operator delete(ptr) to free allocated memory? Then is delete ptr technically equivalent to the following:
T * ptr = new T;
//delete ptr equivalent:
ptr->~T();
::operator delete(static_cast<void *>(ptr));
?
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It is also called general purpose pointer. It is not safe to delete a void pointer in C/C++ because delete needs to call the destructor of whatever object it's destroying, and it is impossible to do that if it doesn't know the type.
delete is used for one single pointer and delete[] is used for deleting an array through a pointer. This might help you to understand better.
The delete operator removes a given property from an object. On successful deletion, it will return true , else false will be returned.
When delete is used to deallocate memory for a C++ class object, the object's destructor is called before the object's memory is deallocated (if the object has a destructor). If the operand to the delete operator is a modifiable l-value, its value is undefined after the object is deleted.
delete ptr will do overload resolution for operator delete, so it may not call the global ::operator delete
But otherwise, yes. The delete operator calls the relevant destructor, if any, and then calls operator delete.
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