delete expression

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Destructs object(s) previously allocated by the new expression and releases obtained memory area

[edit] Syntax

::(optional)    delete    expression (1)
::(optional)    delete [] expression (2)
1) Destroys one non-array object created by a new-expression
2) Destroys an array created by a new[]-expression

[edit] Explanation

For the first (non-array) form, expression must be a pointer or a class with a user-defined conversion function to a pointer, and its value must be either null or pointer to a non-array object created by a new-exression, or a pointer to a base subobject of a non-array object created by a new-exression (if it's anything else, the behavior is undefined).

For the second (array) form, expression must be a null pointer value or a pointer value previously obtained by an array form of new-expression.

If expression is not a null pointer, the delete expression invokes the destructor (if any) for the object that's being destroyed, or for every element of the array being destroyed (proceeding from the last element to the first element of the array).

After that, the delete expression invokes the deallocation function, either operator delete (for the first version of the expression) or operator delete[] (for the second version of the expression).

The deallocation function's name is firstly looked up in the local class type scope and only if the lookup fails, the global namespace is looked up. If :: is present in the delete expression, only the global namespace is looked up. The prototype of the function must look like the following:

void operator delete  (void *ptr);
for the first version
void operator delete[](void *ptr);
for the second version

Both these functions are implicitly declared in each translation unit. Also, implicit implementations are provided by the compiler by default, unless the program has explicitly implemented them. See this for more information.

If expression evaluates to a null pointer value, no destructors are called. The deallocation function may or may not be called (it's implementation-defined), but the default deallocation functions are guaranteed to do nothing when handed a null pointer.

If expression evaluates to a pointer to the base class subobject of the object that was allocated with new, the destructor of the base class must be virtual, otherwise the behavior is undefined.

[edit] Keywords

delete