c++ - 嵌套的 lambda 和可变关键字

Question

考虑以下代码:

void f() {
    int a = 3;
    [=]() {
        [=] () mutable {
            a = 5;
        }();
    }();
}

它在 Clang ( https://godbolt.org/z/IEXotM) 上编译，但不在 GCC (https://godbolt.org/z/xWXFe6) 上编译。 GCC 错误:

<source>: In lambda function:

<source>:5:15: error: assignment of read-only variable 'a'

    5 |             a = 5;

      |             ~~^~~

Compiler returned: 1

根据 https://en.cppreference.com/w/cpp/language/lambda ,

Optional sequence of specifiers.The following specifiers are allowed:

mutable: allows body to modify the parameters captured by copy, and to call their non-const member functions

Clang 在这里的行为似乎是正确的。是这样吗？

Answer 1

如果您从引用的链接底部开始阅读，您可以阅读:

If a nested lambda m2 captures something that is also captured by the immediately enclosing lambda m1, then m2's capture is transformed as follows:

• if the enclosing lambda m1 captures by copy, m2 is capturing the non-static member of m1's closure type, not the original variable or this.

• if the enclosing lambda m1 captures by reference, m2 is capturing the original variable or this.

  #include <iostream>
  
  int main()
  {
  int a = 1, b = 1, c = 1;
  
  auto m1 = [a, &b, &c]() mutable {
      auto m2 = [a, b, &c]() mutable {
          std::cout << a << b << c << '\n';
          a = 4; b = 4; c = 4;
      };
      a = 3; b = 3; c = 3;
      m2();
  };
  
  a = 2; b = 2; c = 2;
  
  m1();                             // calls m2() and prints 123
  std::cout << a << b << c << '\n'; // prints 234
  }
  

因此，您正在从封闭的 lambda 中捕获值并修改它，您也需要使其在封闭的 lambda 中可变。