（转载）警惕“delete void*”

（转载）http://blog.csdn.net/yucan1001/article/details/7001915

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。