# V772\. Calling the 'delete' operator for a void pointer will cause undefined behavior\.

The analyzer detected a possible error that has to do with using the operator delete or operator delete\[\] together with a non\-typed pointer \(void\*\)\. As specified by the C\+\+ standard \(subclause $7\.6\.2\.8/3\), such use of delete results in undefined behavior\.

Consider the following example:

```cpp
class Example
{
  int *buf;

public:
  Example(size_t n = 1024) { buf = new int[n]; }
  ~Example() { delete[] buf; }
};

....
void *ptr = new Example();
....
delete ptr;
....
```

What is dangerous about this code is that the compiler does not actually know the type of the `ptr` pointer\. Therefore, deleting a non\-typed pointer may cause various defects, for example, a memory leak, as the `delete` operator will not call the destructor for the object of `Example` type pointed to by `ptr`\.

If you really mean to use a non\-typed pointer, then you need to cast it to the original type before using `delete` \(`delete[]`\), for example:

```cpp
....
void *ptr = new Example();
....
delete (Example*)ptr;
....
```

Otherwise, it is recommended that you use only typed pointers with `delete` \(`delete[]`\) to avoid errors:

```cpp
....
Example *ptr = new Example();
....
delete ptr;
....
```