警惕“delete void*”
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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 storageconst 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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