临时对象基于范围为 [英] temporary object in range-based for

问题描述

我知道一般来说，基于范围的循环中的一个临时的生命周期被扩展到整个循环（我已经读过C++11：语句的基于范围：range-init生命周期？）。因此，这样的东西一般是这样的：

I know that in general the life time of a temporary in a range-based for loop is extended to the whole loop (I've read C++11: The range-based for statement: "range-init" lifetime?). Therefore doing stuff like this is generally OK:

for (auto &thingy : func_that_returns_eg_a_vector())
  std::cout << thingy;

现在，当我尝试做某些我认为与Qt类似的事情时， QList 容器：

Now I'm stumbling about memory issues when I try to do something I thought to be similar with Qt's QList container:

#include <iostream>
#include <QList>

int main() {
  for (auto i : QList<int>{} << 1 << 2 << 3)
    std::cout << i << std::endl;
  return 0;
}

这里的问题是valgrind显示无效的内存访问 QList 类。然而，修改示例以使列表存储在变量中提供了正确的结果：

The problem here is that valgrind shows invalid memory access somewhere inside the QList class. However, modifying the example so that the list is stored in variable provides a correct result:

#include <iostream>
#include <QList>

int main() {
  auto things = QList<int>{} << 1 << 2 << 3;
  for (auto i : things)
    std::cout << i << std::endl;
  return 0;
}

现在我的问题是：我在第一种情况下做一些事例如未定义的行为（我没有足够的经验阅读C ++标准为了自己回答这个问题）？或者这是一个问题，我如何使用 QList ，或如何实现 QList ？

推荐答案

由于您使用的是C ++ 11，因此您可以使用初始化列表。这将通过valgrind：

Since you're using C++11, you could use initialization list instead. This will pass valgrind:

int main() {
  for (auto i : QList<int>{1, 2, 3})
    std::cout << i << std::endl;
  return 0;
}

---

涉及基于范围的或甚至C ++ 11。下面的代码演示了同样的问题：

---

The problem is not totally related to range-based for or even C++11. The following code demonstrates the same problem:

QList<int>& things = QList<int>() << 1;
things.end();

或：

#include <iostream>

struct S {
    int* x;

    S() { x = NULL; }
    ~S() { delete x; }

    S& foo(int y) {
        x = new int(y);
        return *this;
    }
};

int main() {
    S& things = S().foo(2);
    std::cout << *things.x << std::endl;
    return 0;
}

无效读取是因为表达式 S（）（或 QList< int> {} ）在声明之后被破坏（在C ++ 03和C ++ 11§ 12.2 / 5），因为编译器不知道方法 foo（）（或 operator ）将返回该临时对象。所以你现在指的是释放内存的内容。

The invalid read is because the temporary object from the expression S() (or QList<int>{}) is destructed after the declaration (following C++03 and C++11 §12.2/5), because the compiler has no idea that the method foo() (or operator<<) will return that temporary object. So you are now refering to content of freed memory.

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