C++11 Change `auto` Lambda to a different Lambda?

Question

Say I have the following variable containing a lambda:

auto a = [] { return true; };

And I want a to return false later on. Could I do something along the lines of this?

a = [] { return false; };

This syntax gives me the following errors:

binary '=' : no operator found which takes a right-hand operand of type 
'main::<lambda_a7185966f92d197a64e4878ceff8af4a>' (or there is no acceptable conversion)

IntelliSense: no operator "=" matches these operands
        operand types are: lambda []bool ()->bool = lambda []bool ()->bool

Is there any way to achieve something like this? I would like to change the auto variable to a different lambda. I'm relitively a beginner so I may be missing some knowledge about auto or lambdas. Thanks.

Answer 1

Every lambda expression creates a new unique type, so the type of your first lambda is different from the type of your second (example). Additionally the copy-assignment operator of a lambda is defined as deleted (example) so you're doubly-unable to do this. For a similar effect, you can have a be a std::function object though it'll cost you some performance

std::function<bool()> a = [] { return true; };
a = [] { return false; };

Answer 2

A Lambda may be converted to a function pointer using the unary + operator like so:

+[]{return true;}

so long as the capture group is empty and it doesn't have auto arguments.¹

If you do this, you may assign different lambdas to the same variable as long as the lambdas all have the same signature.

In your case,

auto a = +[]{return true;};
a = +[]{return false;};

^{Live example on Coliru}

would both compile and act as you expect.² You may use the function pointers in the same way you would expect to use a lambda, since both will act as functors.

---

^{1. In C++14, you can declare lambdas with auto as the argument type, like [](auto t){}. These are generic lambdas, and have a templated operator(). Since a function pointer can't represent a templated function, the + trick won't work with generic lambdas.}

^{2. Technically, you don't need the second + operator on the assignment. The lambda would convert to the function pointer type on assignment. I like the consistency, though.}

Answer 3

Each lambda has a different type so you cannot change it. You could use a std::function to hold an arbitrary callable object, that can be changed at will.

std::function <bool ()> a = [] { return true; };
a = [] { return false; };

Answer 4

We may use retrospective call to convert lambda to std::function:

template<typename T>
struct memfun_type
{
  using type = void;
};

template<typename Ret, typename Class, typename... Args>
struct memfun_type<Ret(Class::*)(Args...) const>
{
  using type = std::function<Ret(Args...)>;
};

template<typename F>
typename memfun_type<decltype(&F::operator())>::type
FFL(F const &func)
{ // Function from lambda !
  return func;
}

After that we will be able to do (since 'a' is std::function type now ):

auto a = FFL([] { return false; });
a = FFL([] { return true; });

Answer 5

Since C++17 you can have the std::function template parameter deduced thanks to Class template argument deduction. It even works with capturing lambdas:

int a = 24;

std::function f = [&a] (int p) { return p + a; };
f               = [&a] (int p) { return p - a; };
f               = []   (int p) { return p; };

This comes in handy with more complex signatures and even more with deduced return types.