C++0x lambda 按值捕获总是常量?

问题描述

有没有办法按值捕获，并使捕获的值非常量?我有一个库仿函数，我想捕获 &调用一个非常量但应该是的方法.

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.

以下内容不能编译，但使 foo::operator() const 修复它.

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 对象的运算符 () const.

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

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