"No matching function" for template function accepting a callback

Question

I'm trying to pass a callback to a template function, but GCC gives me

error: no matching function for call to ‘Test::iterate(main(int, char**)::<anonymous struct>&)’

Why doesn't this work? (Also, for reasons beyond my control, I can't use C++11.)

I've also tried naming the struct e.g. myvis and calling test.iterate<myvis>(visitor), but that didn't work either.

#include <deque>
#include <iostream>

class Test {
public:
    std::deque<int> d;

    template <typename C>
    void iterate(C& c) {
        for(std::deque<int>::iterator itr = d.begin(); itr != d.end(); itr++) {
            c(*itr);
        }
    }
};

int main(int argc, char** argv) {
    Test test;
    test.d.push_back(1);
    test.d.push_back(2);
    struct {
        void operator()(int x) {
            std::cout << x << std::endl;
        }
    } visitor;
    test.iterate(visitor);
}

Try defining vistor globally. I think you have a scope issue — Charlie, Aug 20 '12 at 21:05
The answer can be found in [Using local classes with STL algorithms](http://stackoverflow.com/questions/742607/using-local-classes-with-stl-algorithms). The short version is that in C++03 you cannot use a local class in a template. If you define the visitor outside of `main`, as Charlie suggests, it will work — David Rodríguez - dribeas, Aug 20 '12 at 21:06

Grizzly · Accepted Answer · 2012-08-20T21:21:11.370

3

The C++03 standard says the following in §14.3.1.2 [temp.arg.type]:

A local type, a type with no linkage, an unnamed type or a type compounded from any of these types shall not be used as a template-argument for a template type-parameter.

Therefore you need a global named struct instead of a local unnamed one, giving you something like the following:

struct Visitor {
    void operator()(int x) {
        std::cout << x << std::endl;
    }
} visitor;
int main(int argc, char** argv) {
    Test test;
    test.d.push_back(1);
    test.d.push_back(2);
    Visitor visitor;
    test.iterate(visitor);
}

The restrictions on local/unnamed typed have been lifted in c++11, so if those reasons not to use it go away someday your code will be fine as is.

edited Aug 20 '12 at 21:21

answered Aug 20 '12 at 21:06

Grizzly

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... and you have to name your struct. – Karoly Horvath Aug 20 '12 at 21:09
1

I don't see why the restriction on local types was there, but I'm glad they lifted it in C++11. Thanks! – Adam Crume Aug 20 '12 at 21:25
… of course in C++11 you don’t need this convoluted workaround anymore anyway. ;-) – Konrad Rudolph Aug 20 '12 at 21:29
1

@KonradRudolph:... you don't even need to *name* the struct and go through the boilerplate code, just create a lambda. – David Rodríguez - dribeas Aug 20 '12 at 21:40
@KonradRudolph: Oh, sorry... I thought the *convoluted workaround* was having to move the type :) – David Rodríguez - dribeas Aug 20 '12 at 21:41

juanchopanza · Answer 2 · 2012-08-21T05:57:50.347

1

You have two errors.

First of all, you cannot use the local type visitor to instantiate the template member function function iterate (this would work in C++11).

Second, you cannot have an anonymous type here. You need a named struct visitor and you would need to pass an instance of that.

#include <deque>
#include <iostream>

struct visitor {
  void operator()(int x) {
    std::cout << x << std::endl;
  }
};

class Test {
public:
  std::deque<int> d;

  template <typename C>
  void iterate(C& c) {
    for(std::deque<int>::iterator itr = d.begin(); itr != d.end(); itr++) {
      c(*itr);
    }
  }
};

int main() {
  Test test;
  test.d.push_back(1);
  test.d.push_back(2);

  visitor v;
  test.iterate(v);
}

edited Aug 21 '12 at 05:57

answered Aug 20 '12 at 21:13

juanchopanza

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where did the OP try to pass in a temporary? – Grizzly Aug 20 '12 at 21:19
@Grizzly my bad, I am not used to the funny syntax. – juanchopanza Aug 20 '12 at 21:23
visitor isn't a variable, it's a type. I'm not using a temporary. – Adam Crume Aug 20 '12 at 21:23
@AdamCrume you cannot pass a type to the `iterate` method. It needs a value. – juanchopanza Aug 20 '12 at 21:24
@AdamCrume: As written in the question visitor is a variable. – Grizzly Aug 20 '12 at 21:30
@AdamCrume `struct { } someName;` is an instance of an anonymous struct. – juanchopanza Aug 20 '12 at 21:31
Sorry, I got my comment backwards. I meant that visitor in my code is a variable, not a type. – Adam Crume Aug 20 '12 at 21:50

score 1 · Answer 3 · edited Aug 20 '12 at 21:31

1

You could make visitor global and you'll also need to name the struct, say something like Visitor.

#include <deque>
#include <iostream>

struct Visitor {
    void operator()(int x) 
    {
       std::cout << x << std::endl;
    }
};


class Test {
public:
    std::deque<int> d;

    template <typename C>
    void iterate(C& c) {
        for(std::deque<int>::iterator itr = d.begin(); itr != d.end(); itr++) {
            c(*itr);
        }
    }
};

int main(int argc, char** argv) {
    Test test;
    test.d.push_back(1);
    test.d.push_back(2);
    Visitor visitor;
    test.iterate(visitor);
}

edited Aug 20 '12 at 21:31

Grizzly

19,595
4
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answered Aug 20 '12 at 21:17

MartyE

636
3
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Wouldn't it make more sense to keep the variable local (unlike the struct which must be global)? Afterall unnecessary global variables are rarely a good idea for code simplicity and correctness. – Grizzly Aug 20 '12 at 21:23
@Grizzly You are correct, keeping the variable `visitor` local could (and most of the time would) make more sense. Edit made – MartyE Aug 20 '12 at 21:25
I took the libery of actually declaring a local variable, since you seem to have missed that in your edit. – Grizzly Aug 20 '12 at 21:32

"No matching function" for template function accepting a callback

3 Answers3