是以下新的重载泄漏内存？ [英] is the following new overload leaking memory?

问题描述

我遇到了以下代码：

    class a {
    public:

        void *  operator new(size_t l, int nb);
        double  values;
    };
void *a::operator new (size_t l,int n)
{
    return new char[l+ (n>1 ? n - 1 : 0)*sizeof(double)];
}

根据我得到的结果， value：

From what I get it is then used to have an array like structure that start at "values":

double* Val = &(p->a->values) + fColumnNumber;

我的问题是：
是否有内存泄漏？我是非常新的重载new运算符，但我很确定分配的内存没有正确释放。这也意味着我不能在堆栈上创建一个a类？

My question is : is there a memory leak? I am very new to overloading new operator, but I'm pretty sure that the memory allocated is not deallocated properly. Also does that mean I can never create a "a" class on the stack?

感谢

推荐答案

我相信它在技术上生产UB，虽然它是一种形式的UB，可能永远不会产生可见的副作用（它使用 new [] ，但是我相信会匹配 delete - 但对于 char t引起一个可见的问题）。

I believe it technically produces UB as it is, though it's a form of UB that will probably never cause a visible side effect (it's using new [], but I believe that'll get matched up with delete -- but for char, this usually won't cause a visible problem).

IMO，几乎更糟糕的是，它使用一个新的表达式分配什么应该是真正的原始字节而不是对象。如果我这样做，我会这样写：

IMO, it's almost worse that it's using a new expression to allocate what should really be raw bytes instead of objects. If I were doing it, I'd write it like:

void *a::operator new (size_t l,int n)
{
    return ::operator new(l+ (n>1 ? n - 1 : 0)*sizeof(double));
}

您可以将其与：

void a::operator delete(void *block)
{
    ::operator delete(block);
}

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