Default constructor for lambdas

[Vincent Reverdy]

Hello.

What is the reason why lambdas have no default constructor? Is there any technical reason behind that, or is it a pure design decision?

I have a use case where not having a default constructor for lambda is annoying.

Given C++17 class template deduction, I would like to be able to do:

template <class F>
class object
{
    constexpr object(F) {}            // Constructor
    constexpr void execute() {F()();} // Default-construct the lambda, and run it
};

object myobject([](){std::cout<<"Hello world!"<<std::endl;});
myobject.execute();

What do you think? Remarks? Comments?

Thank you very much.
Best,
Vincent Reverdy

[D. B.]

My uninformed guess is that it's because lambdas are primarily intended to be called and passed around as opaque entities, and the fact they have types is a kind of unavoidable baggage, rather than a design goal or something to exploit. But I'm quite likely to be wrong :)

Could you elaborate on where an ability to construct another instance of the templated type would be useful? As written, I can't see why you wouldn't just stash it as a member in the ctor and call it again later. it seems wasteful, constructing a lambda initially, only to discard it and rebuild on each execute(); the one in the ctor is merely a surrogate for type deduction. If the type had mutable or static members, that could cause different behaviour, but then why would you pass such a type into the above pattern? There must be motivations this doesn't illuminate (or maybe it's just too late in the day for me to be thinking about things like this!)

[szollos...@gmail.com]

Hi,

Minor thing - it will only work if all the captures of the lambda are copyable. Even then it's non-trivial sometimes to choose schematics.

Thanks,
-lorro

[Nicol Bolas]

I think this would be much more reasonable as this:

struct HelloWorld { void operator()() {std::cout<<"Hello world!"<<std::endl;}};

object myobject(HelloWorld());

I generally consider globally-scoped lambdas like these to be dubious constructs. There's nothing to capture, so the only thing you're doing is making it slightly more convenient to define a struct with an operator() overload.

That being said, I'm not against the feature, so long as it only applies to lambdas that have no captures. There was a long thread on this forum about applying default constructors to capturing lambdas, and it was generally agreed on that it broke the meaning of the lambda.

[Thiago Macieira]

On sexta-feira, 16 de dezembro de 2016 14:47:45 PST Vincent Reverdy wrote:
> Hello.
>
> What is the reason why lambdas have no default constructor? Is there any
> technical reason behind that, or is it a pure design decision?

What would you initialise the captures-by-reference to?

[i] { ++i; }

has struct equivalent:

struct Lambda
{
int &i;
Lambda(int &i) : i(i) {}
operator()() { ++i; }
};

The above cannot have a default constructor.

--
Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
Software Architect - Intel Open Source Technology Center

[Thiago Macieira]

On sexta-feira, 16 de dezembro de 2016 22:33:51 PST Thiago Macieira wrote:
> [i] { ++i; }

I meant [&i] { ++i; }