我使用三种内存清除方式。他们都安全吗?我可以得到内存泄漏吗?

时间:2013-10-04 13:13:33

标签: c++ visual-c++

我使用“placement new”来分配我的对象。我使用三种内存清除方式。他们都安全吗?我可以得到内存泄漏吗?

#include <iostream>
#include <exception>
#include <vector>
using namespace ::std;

class A{
private:
    double x;
public:
    A() : x(0) { cout << "A class; ptr: " << this << " created." << endl; } 
    ~A() { cout << "A class; ptr: " << this << " destroyed." << endl; }
};

int main(int argc, char* argv[])
try{
    // 1. Creating of object in the necessary memory address

    static_assert(sizeof(char) == 1, "Unexpected size of char.");
    int x = -1; // Variants of memory clearing
    while (x < 0 || x > 2) {
        cout << "Variant (0,1,2): ";
        cin >> x;
    }
    char* p = new char[sizeof(A)]; // some memory area...

    A* a = new(p)A(); // Place my object in the 'p' address.

    // Here is my basic work to do...

    // Now I must to free my memory:
    if(!x){ // First variant
        delete a;           
    }
    else if (x == 1){ // Second variant
        delete reinterpret_cast<A*>(p); 
    }
    else if (x == 2){ // Third variant
        a->~A();        
        delete[] p; 
    }
    else{
        throw runtime_error("Invalid variant!");
    }
    a = nullptr;
    p = nullptr;

    cout << endl;   
}
catch(exception& e){
    cerr << e.what() << endl;
    return 1;
}
catch(...){
    cerr << "Unknown exception." << endl;
    return 2;
}

谢谢。

2 个答案:

答案 0 :(得分:3)

具有delete[]和显式析构函数调用的变体是正确的,因为它反映了您如何分配/构造它:

char* p = new char[sizeof(A)];
A* a = new(p)A();
...
a->~A();        
delete[] p; 

但如果您没有充分的理由使用新的展示位置,请考虑简单明了:

A* a = new A();
...
delete a;

虽然delete应该为每个newdelete[]调用new[],但是因为您分配了char s的数组,所以第二个选项看起来似乎不合理,但仍然合法(只要您确定内存块的大小确实等于sizeof(A)并且存在类型为A的有效对象在这个数组内):

char* p = new char[sizeof(A)];
delete reinterpret_cast<A*>(p);

另请注意,以下行完全没用:

static_assert(sizeof(char) == 1, "Unexpected size of char.");

因为标准保证sizeof(char)总是返回1。

答案 1 :(得分:1)

第三种变体是删除对象并清除已分配内存的正确方法。