Function default parameters are ignored

Question

For this simplified piece of code I'm getting following error:

error: too few arguments to function std::cout << f();

int g(int a = 2, int b = 1)
{
    return a + b;
}

template<class Func>
void generic(Func f)
{
    std::cout << f();
}

int main()
{
    generic(g);
}

I cannot clue the reason for why the default parameters of function f are not passing into a function generic. It behaves like f doesn't have any default parameters ...

What's wrong there?

How do I forward default parameters correctly?

I'm guessing that the time the function pointer is passed, it already selected an overload for the function that needs 2 parameters. — Peter Lenkefi
– Peter Lenkefi, Commented Sep 13, 2017 at 19:41
Language lawyer alert… a and b are function parameters. These parameters have default arguments of 2 and 1, rsp. So a is a parameter, and the value of a at run time is the argument. — Arne Vogel
– Arne Vogel, Commented Sep 14, 2017 at 9:56

Barry · Accepted Answer · 2017-09-13 19:41:55Z

22

g may have default arguments, but the type of &g is still int(*)(int, int), which is not a type that can be called with no arguments. Within generic, we can't differentiate that - we've already lost the context about default arguments.

You can just wrap g in a lambda to preserve the context:

generic([]{ return g(); });

answered Sep 13, 2017 at 19:41

Barry

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6 Comments

463035818_is_not_an_ai Over a year ago

is there any possibility (without bind) to get a int(*)(int) or int(*)() from g ?

Rakete1111 Over a year ago

Maybe soon we will able to do this generic([]() => g());, right? :P (thanks)

Barry Over a year ago

@tobi303 You can't do it with bind at all - same problem as here. Has to be a lambda.

ampawd Over a year ago

hmm that's interesting, it seems to be working that way. But how does lambda preserve the context ?

Barry Over a year ago

@ampawd Because g() is a valid expression, you're invoking it directly by name. It's the name lookup via g that gets you the default arguments, not through the type of g

|

Peter Mortensen · Accepted Answer · 2017-09-13 22:37:49Z

7

I think the error message to this code quite nicely demonstrates why this isn't possible:

int g(int a=0,int b=0){return 0;}

typedef int (*F1)(int);

int main() {
    F1 x = g;
}

error: cannot initialize a variable of type 'F1' (aka 'int (*)(int)') with
an lvalue of type 'int (int, int)': different number of parameters (1 vs 2)
    F1 x = g;
       ^   ~

Even with default parameters, the type of g is still

int (*) (int,int)

and that's what gets deduced when you instantiate the template.

If for some reason you cannot use C++11 or later (i.e. no lambdas, see Barry's answer) and you don't mind a little boilerplate, then you can use a function object:

#include <iostream>

struct g_functor {
    int operator()(int a=0,int b=0){ return a;}
};

template <typename T> void foo(T t) { t(); }

int main() { foo(g_functor()); }

Note that you have to create an instance of g_functor to pass it as a parameter.

edited Sep 13, 2017 at 22:37

Peter Mortensen

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answered Sep 13, 2017 at 19:52

463035818_is_not_an_ai

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3 Comments

Barry Over a year ago

If you're going to go boilerplate, you should make g a function object instead - so you can still do generic(g)

ampawd Over a year ago

yeah, nice solution with function object as well

463035818_is_not_an_ai Over a year ago

@ampawd function objects are just the poor mans lambdas (or lambdas are just function objects with some fancier syntax). I was just curious to see how one would do it in C++98. For me this always helps to better understand the C++11 features as most of them were invented to overcome C++98 shortcomings

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