通过引用捕获C ++ Lambda

Question

As far I know, after running the following code myString should have got the value "Inside lambda3.secondLambda". Instead it will still have the value "Initialized".

std::string myString("Initialized");
auto lambda3 = [&]() { [&myString]() { myString = "Inside lambda3.secondLambda"; }; };

The first lambda captures all variables by reference and the sencond captures only myString by reference. Why does it not behave as I have expected?

Answer 1

By specifying the body of the lambda, you specify code that would be called when the lambda's operator() is called.

So you need to actually call the lambdas, otherwise it's simply code inside the definition of functions which never get called. You can do:

auto lambda3 = [&] {
  auto l = [&myString] {
    myString = "Inside lambda3.secondLambda";
  };
  l();
};
lambda3();

To take it one step further, you can return the inner lambda and then execute it like this:

auto lambda3 = [&] {
  return [&myString] {
    myString = "Inside lambda3.secondLambda";
  };
};
lambda3()();

Answer 2

You have defined a closure, but not actually run the function. You must actually run it like lambda3() if you want to execute its body.

However, you have nested lambdas, so you need to run what's in the lambda as well. It's not clear why you are doing this. Your inner lambda is neither assigned inside the outer lambda's scope (so you can run it inside the lambda using the same syntax), or returned (so the caller of this lambda can run what it returns).

If you returned it, you could do lambda3()() :

auto lambda3 = [&]() { return [&myString]() { myString = "Inside    
     lambda3.secondLambda"; }; };

Note: only the word return was added to allow lambda3()() to work.

In this particular example, you are better served with just:

auto lambda3 = [&myString]() { 
      myString = "Inside lambda3.secondLambda"; };