c++ - 我使用三种内存清除变体。他们都安全吗？我可以得到内存泄漏吗？

Question

我使用“placement new”来分配我的对象。我使用三种内存清理变体。他们都安全吗？我会发生内存泄漏吗？

#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。

Answer 1

带有 delete[] 和显式析构函数调用的变体是正确的，因为它是您如何分配/构造它的镜像:

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

但是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;

尽管应该为每个 new 调用 delete 并为每个 new[] 调用 delete[]，因为你分配了一个 char 数组，所以第二个选项似乎不太合理，但仍然合法(只要你确定内存块的大小确实等于 sizeof(A) 并且此数组中存在类型为 A 的有效对象:

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。