specify default value of std::function

Question

void func(const std::function<void()>& f = empty)
{
    if(f)
        f();
}

what is the 'empty' should be? I use [](){} . But technically, that is not empty, the f() will execute.

101010 · Accepted Answer · 2015-11-20T08:55:56.807

76

void func(const std::function<void()>& f = {}) {
    if(f) f();
}

LIVE DEMO

edited Nov 20 '15 at 08:55

answered Nov 20 '15 at 08:44

101010

41,839
11
94
168

@suzuiyue What do you mean? – 101010 Aug 17 '17 at 13:38
I do not see bool cast or ! operator overload in [doc](http://en.cppreference.com/w/cpp/utility/functional/function). So why if(f) or if (!f) works? – suzuiyue Aug 18 '17 at 03:01
@suzuiyue I am just getting into std::funtion now, but I would assume it's because it was created as a C++ alternative to C-style function pointers, where you check whether or not the fp is null, and f would be a function pointer to the unnamed function created by the lambda. Someone can correct me here if I'm wrong. – Sir Rogers Nov 30 '17 at 11:24
4

@suzuiyue http://en.cppreference.com/w/cpp/utility/functional/function/operator_bool – 101010 Nov 30 '17 at 11:33

score 22 · Answer 2 · edited Nov 20 '15 at 13:54

22

const std::function<void()>& f = nullptr

or

const std::function<void()>& f = std::function<void()>()

edited Nov 20 '15 at 13:54

Soner Gönül

97,193
102
206
364

answered Nov 20 '15 at 08:42

nh_

2,211
14
24

2

Yes, and the default `std::function` is throwing some exception (I forgot which one).... – Basile Starynkevitch Nov 20 '15 at 08:44
@BasileStarynkevitch but OP's code checks the function to see if it's empty before calling. Or is that not what you meant? – TartanLlama Nov 20 '15 at 08:45
2

@BasileStarynkevitch:- Looking for this one: `1/ An exception of type bad_function_call is thrown by function::operator() (20.8.11.2.4) when the function wrapper object has no target.` ? – Rahul Tripathi Nov 20 '15 at 08:45
[nullptr reference is not valid](https://stackoverflow.com/a/4364586/5393174) – kgbook Aug 29 '20 at 07:11
1

@kgbook not sure that it's what you meant, but `const std::function& f = nullptr` is totally valid. `function& operator=( std::nullptr_t ) noexcept` is defined and drops the current target, making the function empty. – Adrian B. Apr 21 '22 at 16:58

score 10 · Answer 3 · answered Nov 20 '15 at 09:24

I would use a nullptr for the empty case. It can be used because you never use full function objects but only pointers to them, so func(f1); will be in fact func(&f1) => you will always pass pointers to func, so nullptr is IMHO the best candidate.

This compiles and run:

void func(const std::function<void()>& f = nullptr)
{
    if(f)
        f();
}

The alternative of using a default function would be:

void func(const std::function<void()>& f = std::function<void()>()) {
    try {
        f();
    }
    catch(std::bad_function_call e) {
    }
}

using exception catching. The choice between the 2 mainly depends on whether you expect the empty case to occur; almost never: go with exception vs. sometimes: test before call.

As question mentions if (f), use nullptr

score 3 · Answer 4 · answered Nov 20 '15 at 08:43

You can try like this:

const std::function<void()> & f = std::function<void()>()

As Basile added, the standard says:

§ 20.8.11.1 Class bad_function_call [func.wrap.badcall]

1/ An exception of type bad_function_call is thrown by function::operator() (20.8.11.2.4) when the function wrapper object has no target.

specify default value of std::function

4 Answers4