C ++ 0x lambda按值捕获总是const?

2022-01-23 00:00:00 lambda constants c++ c++11

有什么方法可以按值捕获，并使捕获的值非常量?我有一个库函子，我想捕获 &调用一个非 const 但应该是的方法.

Is there any way to capture by value, and make the captured value non-const? I have a library functor that I would like to capture & call a method that is non-const but should be.

以下内容无法编译，但将 foo::operator() const 修复.

The following doesn't compile but making foo::operator() const fixes it.

struct foo
{
  bool operator () ( const bool & a )
  {
    return a;
  }
};


int _tmain(int argc, _TCHAR* argv[])
{
  foo afoo;

  auto bar = [=] () -> bool
    {
      afoo(true);
    };

  return 0;
}

推荐答案

使用可变的.


auto bar = [=] () mutable -> bool ....

如果没有 mutable，您将声明 lambda 对象的 operator () const.

Without mutable you are declaring the operator () of the lambda object const.

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