在被调用函数内部更改函数指针（std :: function）是否安全？ [英] Is it safe to change a function pointer (std::function) inside a called function?

问题描述

我有一个 std :: function 指向一个函数。在此函数内，我将指针更改为另一个函数。

I have a std::function pointing to a function. Inside this function I change the pointer to another function.

std::function<void()> fun;

void foo() {
    std::cout << "foo\n";
}

void bar() {
    std::cout << "bar\n";
    fun = foo;
}

int main() {
    fun = bar;
    fun();
    fun();
}

我看不到任何问题，而且效果很好（请参阅此处），但是我不确定这样做是否合法。我有什么想念的吗？也许在c ++标准草案中（我检查很快，但到目前为止什么都没看到）。

I can't see any problem and it works just fine (see here), however I'm not sure if this is legal to do so. Is there anything I am missing? Maybe in the c++ standard draft (I checked quickly but didn't see anything so far).

推荐答案

这是合法的函数

当您为目标分配或构造一个 std :: function 时，它会创建一个复制目标。在为 std :: function 分配功能的情况下，这实际上将功能指针存储为目标对象。

When you assign or construct a std::function with a target, it creates a copy of the target. In the case of assigning a function to the std::function, this in effect stores a function pointer as the target object.

当您调用 operator（）时，需要返回如果您使用参数调用目标时会发生的情况。

When you invoke operator(), it is required to return what would happen if you invoked that the target with the arguments.

在函数对象副本的主体中，如果您重新分配给<$，则该副本作为副本存储在 std :: function 中c $ c> std :: function ，这会破坏旧的目标函数对象。

Within that "body" of the copy of the function object stored as a copy in the std::function, if you reassign to the std::function, this destroys the old target function object.

销毁函数指针不会影响在指向的函数中执行的代码。

Destroying a function pointer has no impact on the validity of code executed within the function pointed to.

但是，如果您存储了函数 objects （lambda，手动值，其他 std :: function s， std :: bind 等），在分配时，您会遇到通常的运行规则 this 被销毁时的类中的方法。简而言之，您将无法执行依赖于实例的本地状态的任何操作。

However, if you had stored function objects (lambdas, manual ones, other std::functions, std::bind, etc), at the point of assignment you'd run into the usual rules of running a method in a class when this is destroyed. In short, you would no longer be able to do anything that relied on "local state" of your instance.

std::function<void()> fun;
struct bob {
  std::string name;
  bob* next = 0;
  void operator()() const {
    std::cout << name << "\n";
    if (next) fun = *next;
    // undefined behavior:
    // std::cout << name << "\n";
  }
};
bob foo = {"foo"};
bob bar = {"bar", &foo};

int main() {
  fun = bar;
  fun();
  fun();
}

实时示例。

因此您可以看到，这可能很脆弱。

So as you can see, this can be fragile.

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