c++ - 在 C++ lambdas 中捕获内部构造函数的类成员

Question

假设我们有 A 类:

class A
{
    double x;
    double y;
    A(double, double);
    function <double(void)> F;
};

以及以下构造函数:

A::A(double a, double b)
{
    x = a;
    y = b;
    F = [this]() { return x + y; };
}

为什么上面的构造函数可以工作，而下面的构造函数会导致编译错误:member A::x is not a variable？ (y 的相同错误。)

A::A(double a, double b)
{
    x = a;
    y = b;
    F = [x,y]() { return x + y; };
}

看来我只能捕获this 而不能捕获类成员。这是为什么？

Answer 1

来自 cppreference – Lambda expressions :

Class members cannot be captured explicitly by a capture without initializer (as mentioned above, only variables are permitted in the capture list):

class S {
  int x = 0;
  void f() {
    int i = 0;
//  auto l1 = [i, x]{ use(i, x); };    // error: x is not a variable
    auto l2 = [i, x=x]{ use(i, x); };  // OK, copy capture
    i = 1; x = 1; l2(); // calls use(0,0)
    auto l3 = [i, &x=x]{ use(i, x); }; // OK, reference capture
    i = 2; x = 2; l3(); // calls use(1,2)
  }
};