Function Objects Set in C++

Question

So basically I want to create a set of Function objects. In python if we do:

def func():
    print "a"

a = func
b = func
fset = set()
fset.insert(a)
fset.insert(b)

In this case fset will have only one function since both a and b are same in python. But in C++, if I create function objects for same function both a and b will be two different objects of a set. Is there any way that two objects of same function be same?

In C++:

void func(){
    cout << "a";
}

function<void()> a = bind(func);
function<void()> b = bind(func);

Now I want if a or its pointer is already present in the set, b should not be added.

a and b are essentially two pointers to the same object, not two instances of the same class. You can have std::set<std::function<void()>*>, or possibly a smart pointer - it'll behave in a similar manner. — Igor Tandetnik
– Igor Tandetnik, Commented Nov 23, 2018 at 5:22
But in this case also every time creating a pointer of function<void()> a = function(); function<void()> b = function(); Both a and b will still have different pointers right? — 250
– 250, Commented Nov 23, 2018 at 5:23
A set, by definition, contains values distinct from one another. It's not clear what you're talking about here, when you suggest that you are putting the "same function" into a set as two different objects. Perhaps you should back up your question with some C++ code to illustrate your problem. — paddy
– paddy, Commented Nov 23, 2018 at 5:25
You say "creating a pointer", but your code doesn't actually create any pointers. It's also not clear what you mean by function() - that doesn't appear syntactically valid and likely won't compile. — Igor Tandetnik
– Igor Tandetnik, Commented Nov 23, 2018 at 5:25
@IgorTandetnik they are talking about your comment which stores std::function<void()>* — paddy
– paddy, Commented Nov 23, 2018 at 5:26

JeJo · Accepted Answer · 2018-11-23 05:46:04Z

1

If you have only void functions(or all functions have the same signature), use simply C type function pointers as std::sets's template type.

This will work and as a plus no type erasure overheads of std::function.

void func() {}
void func2() {}

using fPtrType = void(*)(); // convenience type  

int main()
{
    std::set<fPtrType> fset;
    fPtrType a = func;
    fPtrType b = func;
    fset.emplace(a);
    fset.emplace(b);
    fset.emplace(func2);

    std::cout << fset.size();  // prints 2
    return 0;
}

answered Nov 23, 2018 at 5:46

JeJo

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2 Comments

250 Over a year ago

Will this also work if both a and b have a different scope?

JeJo Over a year ago

@250 Did you mean this.

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