The ‘Type of Lambda’

In my first post about functional programming styles I sort of skipped over what exactly a lambda expression ‘becomes’ when it’s evaluated. As this is generally not that well documented I thought it would be worth talking about it briefly.

The Standard (N3242) starts out by stating what most people would assume at clause

“Lambda expressions provide a concise way to create simple function objects.”

So we know that the result of evaluating a lambda expression is a function object. This is called the ‘closure object’ and is a prvalue:

“The evaluation of a lambda-expression results in a prvalue temporary. This temporary is called the closure object.” (

As a reminder, ‘prvalue’ stands for pure rvalue. For all practical purposes a prvalue IS an rvalue, it’s just that the standard is being exact here by differentiating two different kinds of rvalues, i.e. xvalues and prvalues.

That’s because there is a subtle difference between an object “near the end of its lifetime” (3.10.1) and something that, for example, ONLY appears on the right hand side of an expression. As there can be no question that the latter is actually an rvalue, the standard calls this a pure rvalue or prvalue:

“An xvalue (an “eXpiring” value) also refers to an object, usually near the end of its lifetime (so that its resources may be moved, for example). […]

— An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment expression) is an xvalue, a temporary object (12.2) or subobject thereof, or a value that is not associated with an object.

— A prvalue (“pure” rvalue) is an rvalue that is not an xvalue. [ Example: The result of calling a function whose return type is not a reference is a prvalue. The value of a literal […] or true is also a prvalue. —end example ]” (3.10.1)

The intuition behind an xvalue is that that it could seen as BOTH an lvalue and an rvalue in context. Consider this snippet of code:


Our object foo is of automatic storage duration which means that it will be deleted at the end of the block. This makes it safe to ‘move’ from foo, i.e. to convert it to an rvalue using std::move. However, our object has a name, ‘foo’, which makes it an lvalue (remember that the rule of thumb is that anything that has a name is an lvalue). It is only because it is nearing the end of its lifetime that we can use it like an rvalue (provided of course we do not access/modify it after having treated it like an rvalue). Thus, foo is an lvalue that can be used like an rvalue, which the standard calls an xvalue. This is expressed in the following diagram found at 3.10.1:


The gist of this is that lambda expressions return temporary objects in the same way that objects are returned by value from functions, which makes them pure rvalues.

Now that we know what kind of thing is being returned by lambda expression, we have to look at its type, i.e. what exactly we are getting. The standard tells us that this type is unique as well as implementation defined, so let’s call it ‘type lambda’ for now (Although if you were speaking standardese you would call this the ‘closure type’).

“The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type — called the closure type — whose properties are described below. […] An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program [..]” (

The last line of this quote is a promise that there are certain things about type lambda that will always be the same and this is what we can exploit when using it. The most important of those things for the purposes of this discussion is that type lambda will have an operator () that matches our definition in the lambda expression:

“The closure type for a lambda-expression has a public inline function call operator (13.5.4) whose parameters and return type are described by the lambda-expression’s parameter-declaration-clause and trailing return-type respectively. This function call operator is declared const (9.3.1) if and only if the lambda expression’s parameter-declaration-clause is not followed by mutable. It is neither virtual nor declared volatile.“ (

This promise is important because it tells us how we can use a type lambda. Let’s assume we are dealing with a simple lambda expression that simply adds its two integer arguments and returns the result by value.

First of all, you could simply use auto like this:


The expression statement ‘X(11, 22)’ is guaranteed to work because of the promise the standard makes, regardless of what the compiler deduces auto to be.

This also means that we can use a std::function of the appropriate type here instead of auto:


Similarly, if you were writing a function that takes a lambda and then applies it to some arguments, simply use templates and then call the overloaded operator:


As an aside, this also gives you a compile-time check whether the lambda is of the correct type which will help to identify errors as early as possible. This is because the compiler knows exactly what operator () will look like for the type you are trying to pass to the function.

I hope this helps clarify some things. As always, thx for reading 🙂