c++ - 捕获 const this

Question

现在我有一个带有 lambda 的对象函数，为了使用我必须的成员函数和变量(或者当然捕获所有......):

void MyClass::MyFunc() {

    auto myLambda = [this](){...};
}

有没有办法明确声明捕获 const this ？我知道我可以:

void MyClass::MyFunc() {
    MyClass const* const_my_class = this;
    auto myLambda = [const_my_class](){...};
}

谢谢。

Answer 1

根据标准 (N3485) 中的§5.1.2，lambda-capture 的定义是:

lambda-capture:    capture-default    capture-list    capture-default , capture-listcapture-default:    &    =capture-list:    capture ... opt    capture-list , capture ... optcapture:    identifier    & identifier    this

So, you only can have =, &, this, identifier, & identifier in the capture list. You can not have expressions, for example casting this to a const.

Some simple expressions in the capture list in higher versions (-std=c++1y) is avaiable, for example:

auto myLambda = [self = static_cast<MyClass const*>(this)](){

    // Use `self` instead of `this` which is `const`

};

当然，不像捕获this那样可以像访问局部变量一样访问成员。