为什么我不能更改由 lambda 函数中的副本捕获的变量的值？

Question

I'm reading the C++ Programming Language by B. Stroustrup in its section 11.4.3.4 "mutable Lambdas", which says the following:

Usually, we don't want to modify the state of the function object (the closure), so by default we can't. That is, the operator()() for the generated function object (§11.4.1) is a const member function. In the unlikely event that we want to modify the state (as opposed to modifying the state of some variable captured by reference; §11.4.3), we can declare the lambda mutable .

I don't understand why the default for the operator()() is const when the variable is captured by value. What's the rational for this? What could go wrong when I change the value of a variable, which is copied into the function object?

Answer 1

One can think of lambdas as classes with operator()() , which by default is defined as const . That is, it cannot change the state of the object. Consequently, the lambda will behave as a regular function and produce the same result every time it is called. If instead, we declare the lambda as mutable, it is possible for the lambda to modify the internal state of the object, and provide a different result for different calls depending on that state. This is not very intuitive and therefore discouraged.

For example, with mutable lambda, this can happen:

#include <iostream>

int main()
{
  int n = 0;
  auto lam = [=]() mutable {
    n += 1;
    return n;
  };

  std::cout << lam() << "\n";  // Prints 1
  std::cout << n << "\n";      // Prints 0
  std::cout << lam() << "\n";  // Prints 2
  std::cout << n << "\n";      // Prints 0
}

Answer 2

It is easier to reason about const data.

By defaulting const, brief lamndas are easier to reason about. If you want mutability you can ask for it.

Many function objects in std are copied around; const objects that are copied have simpler state to track.