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
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