Question

我有一个类似于下面的c ++代码。这将导致Base::add方法和total = &(*total + *to_add);中的内存泄漏。我该如何解决这个问题？

#include <iostream>
#include <string>
class Base
{
    public:
        int n;
        Base(int input) : n(input) {}
        Base(const Base& input)
        {
            n = input.n;
        }
        Base& add(Base &other, bool new_obj=true)
        {
            Base *self;
            if (new_obj) {
                self = new Base(other);
            } else {
                self = this;
            }
            self->n += other.n;
            return *self;
        }
        Base& operator+=(Base &other)
        {
            return this->add(other, false);
        }
};
Base& operator+(Base &self, Base &other)
{
    return self.add(other);
}
class A : public Base
{
    using Base::Base;
    std::string print() {
        return "Class A method_a";
    }
};

class B : public Base
{
    using Base::Base;
    std::string print() {
        return "Class B method_b";
    }
};

int main()
{
    Base *total = new Base(0);
    for (int i=0; i<5; i++) {
        Base *to_add = new A(i);
        total = &(*total + *to_add);
    }
    for (int i=0; i<9; i++) {
        Base *to_add = new B(i);
        total = &(*total + *to_add);
    }
    return 0;
}

Answer 1

C ++不是Java，你应该按价值回归。从您的示例中不清楚派生类的用途是什么，所以我假设您并不真正需要它们：

#include <iostream>
class Base
{
    public:
        int n;
        Base(int input) : n(input) {}
        Base(const Base& input) = default;
};

Base& operator+=(Base &x, const Base &y)
{
    x.n += y.n;
    return x;
}

Base operator+(const Base &x, const Base &y)
{
    return Base(x.n + y.n);
}

int main()
{
    Base total(0);
    for (int i=0; i<5; i++) {
        total += Base(i); // way 1
    }
    for (int i=0; i<9; i++) {
        total = total + Base(i); // way 2
    }
    return 0;
}

Answer 2

total = &(*total + *to_add);

你为什么要对自己这样做？没有total = (total + to_add)为你工作吗？

在我看来，好像你在这里试图做出相互矛盾的事情......

Answer 3

看起来您的Base :: add函数很简单，您可以按如下方式修改它：

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方法返回自身副本时C ++内存泄漏

3 个答案: