警惕“delete void*”

对一个void*类型指针进行delete操作会出错，除非指针所指的内容是简单类型内容，因为这个操作只会释放内存，而不会执行析构函数

下面是一个代码示例：

//:BadVoidPointerDeletion.cpp
#include <iostream>
using namespace std;

class Object
{
	void* data; // Some storage
	const int size;
	const char id;
public:
	Object(int sz, char c) : size(sz), id(c) 
	{
		data = new char[size];
		cout << "Constructing object " << id 
			<< ", size = " << size << endl;
	}
	~Object() 
	{ 
		cout << "Destructing object " << id << endl;
		delete []data; // OK, just releases storage,
		// no destructor calls are necessary
	}
};

int main() 
{
	Object* a = new Object(40, 'a');
	delete a;
	void* b = new Object(40, 'b');
	delete b; //会释放Object对象的内存，但不会释放data所指向的内存，即不会执行析构函数
}

运行结果：

Constructing object a, size = 40
Destructing object a
Constructing object b, size = 40

如果在程序中发现了内存丢失，应该搜索所有的delete语句并检查被删除指针的类型，如果是void*类型，则可能发现了引起内存丢失的某个原因