No matching function call error with usage of, lambda expressions and std::function.

I am trying to use this lambda expression in my code. And looks like something's wrong with definition of functor or lambda.

I believe the lambda expression is true. but I can't satisfy the prototype of function I defined.

Function definition:

    template<typename E, typename container> 

void for_each(Iterator<E,container> begin, std::function<void(E&)> fun)

{

    do // does the given unary functor to elements 

    {  //from starting iterator 'til the end.

        fun(*begin);

        begin = begin.next();

    }while(begin.hasNext());



    fun(*begin);

}

And caller:

  for_each(c.iterator(), [&](E& e){add(e);});

I except achieve this function call with lambda expression. But compiler says "error: no matching function for call to .."

asked Dec 28 '18 at 23:16

drh0use

225

add a comment |

I am trying to use this lambda expression in my code. And looks like something's wrong with definition of functor or lambda.

I believe the lambda expression is true. but I can't satisfy the prototype of function I defined.

Function definition:

    template<typename E, typename container> 

void for_each(Iterator<E,container> begin, std::function<void(E&)> fun)

{

    do // does the given unary functor to elements 

    {  //from starting iterator 'til the end.

        fun(*begin);

        begin = begin.next();

    }while(begin.hasNext());



    fun(*begin);

}

And caller:

  for_each(c.iterator(), [&](E& e){add(e);});

I except achieve this function call with lambda expression. But compiler says "error: no matching function for call to .."

asked Dec 28 '18 at 23:16

drh0use

225

add a comment |

I am trying to use this lambda expression in my code. And looks like something's wrong with definition of functor or lambda.

I believe the lambda expression is true. but I can't satisfy the prototype of function I defined.

Function definition:

    template<typename E, typename container> 

void for_each(Iterator<E,container> begin, std::function<void(E&)> fun)

{

    do // does the given unary functor to elements 

    {  //from starting iterator 'til the end.

        fun(*begin);

        begin = begin.next();

    }while(begin.hasNext());



    fun(*begin);

}

And caller:

  for_each(c.iterator(), [&](E& e){add(e);});

I except achieve this function call with lambda expression. But compiler says "error: no matching function for call to .."

asked Dec 28 '18 at 23:16

drh0use

225

I am trying to use this lambda expression in my code. And looks like something's wrong with definition of functor or lambda.

I believe the lambda expression is true. but I can't satisfy the prototype of function I defined.

Function definition:

    template<typename E, typename container> 

void for_each(Iterator<E,container> begin, std::function<void(E&)> fun)

{

    do // does the given unary functor to elements 

    {  //from starting iterator 'til the end.

        fun(*begin);

        begin = begin.next();

    }while(begin.hasNext());



    fun(*begin);

}

And caller:

  for_each(c.iterator(), [&](E& e){add(e);});

I except achieve this function call with lambda expression. But compiler says "error: no matching function for call to .."

c++ c++11 lambda

asked Dec 28 '18 at 23:16

drh0use

225

asked Dec 28 '18 at 23:16

drh0use

225

asked Dec 28 '18 at 23:16

drh0use

225

asked Dec 28 '18 at 23:16

drh0use

225

asked Dec 28 '18 at 23:16

drh0use

225

add a comment |

2 Answers
2

active

oldest

votes

lambda is not std::function, so deduction fails.

You might simply pass generic Functor:

template<typename E, typename container, typename F> 

void for_each(Iterator<E, container> begin, F&& fun)

{

    // ...

}

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

add a comment |

Issues with the code and Possible solution approaches below:

1. A Lambda expression is not std::function, it can only be stored in an object of type std::function.

Approach: (With your existing signature):

std::function<void()> f = () -> void { // do something };

for_each(vec.begin(), vec.end(), f);

2. Better solution would be to use a Callable as a parameter (Go to end of the answer).

3. If you want to pull up your own version of for_each, why not follow the implementation that is consistent with other algorithms or for that matter with the implementation of for_each in the algorithm header.

Does all you want and is also consistent with other standard algorithms: en.cppreference.com/algorithms/for_each:

    template<class InputIt, class UnaryFunction>

    constexpr UnaryFunction for_each(InputIt first, InputIt last, UnaryFunction f)

    {

      for (; first != last; ++first) {

        f(*first);

      }

      return f;

    }

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

lambda is not std::function, so deduction fails.

You might simply pass generic Functor:

template<typename E, typename container, typename F> 

void for_each(Iterator<E, container> begin, F&& fun)

{

    // ...

}

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

add a comment |

lambda is not std::function, so deduction fails.

You might simply pass generic Functor:

template<typename E, typename container, typename F> 

void for_each(Iterator<E, container> begin, F&& fun)

{

    // ...

}

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

add a comment |

lambda is not std::function, so deduction fails.

You might simply pass generic Functor:

template<typename E, typename container, typename F> 

void for_each(Iterator<E, container> begin, F&& fun)

{

    // ...

}

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

lambda is not std::function, so deduction fails.

You might simply pass generic Functor:

template<typename E, typename container, typename F> 

void for_each(Iterator<E, container> begin, F&& fun)

{

    // ...

}

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

answered Dec 28 '18 at 23:18

Jarod42

114k12101182

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

add a comment |

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

Ty, it worked. But I was trying to imitate this link example. What part did I miss?

– drh0use
Dec 28 '18 at 23:23

Issue is with deduction, you might call it successfully for_each<E>(c.iterator(), [&](E& e){add(e);}); with your version. lambda can convert to std::function, but it is unrelated type.

– Jarod42
Dec 28 '18 at 23:28

A curious decision to pass the function-object by reference instead of by value, like the standard-library does. If indirection is actually wanted, that's what std::ref() is for.

– Deduplicator
Dec 29 '18 at 16:22

@Deduplicator: pre-c++11, there are no forward references. now interfaces use them, for example std::invoke, std::thread, ... (which still might use std::reference_wrapper).

– Jarod42
Dec 29 '18 at 17:01

add a comment |

Issues with the code and Possible solution approaches below:

1. A Lambda expression is not std::function, it can only be stored in an object of type std::function.

Approach: (With your existing signature):

std::function<void()> f = () -> void { // do something };

for_each(vec.begin(), vec.end(), f);

2. Better solution would be to use a Callable as a parameter (Go to end of the answer).

Does all you want and is also consistent with other standard algorithms: en.cppreference.com/algorithms/for_each:

    template<class InputIt, class UnaryFunction>

    constexpr UnaryFunction for_each(InputIt first, InputIt last, UnaryFunction f)

    {

      for (; first != last; ++first) {

        f(*first);

      }

      return f;

    }

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

add a comment |

Issues with the code and Possible solution approaches below:

1. A Lambda expression is not std::function, it can only be stored in an object of type std::function.

Approach: (With your existing signature):

std::function<void()> f = () -> void { // do something };

for_each(vec.begin(), vec.end(), f);

2. Better solution would be to use a Callable as a parameter (Go to end of the answer).

Does all you want and is also consistent with other standard algorithms: en.cppreference.com/algorithms/for_each:

    template<class InputIt, class UnaryFunction>

    constexpr UnaryFunction for_each(InputIt first, InputIt last, UnaryFunction f)

    {

      for (; first != last; ++first) {

        f(*first);

      }

      return f;

    }

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

add a comment |

Issues with the code and Possible solution approaches below:

1. A Lambda expression is not std::function, it can only be stored in an object of type std::function.

Approach: (With your existing signature):

std::function<void()> f = () -> void { // do something };

for_each(vec.begin(), vec.end(), f);

2. Better solution would be to use a Callable as a parameter (Go to end of the answer).

Does all you want and is also consistent with other standard algorithms: en.cppreference.com/algorithms/for_each:

    template<class InputIt, class UnaryFunction>

    constexpr UnaryFunction for_each(InputIt first, InputIt last, UnaryFunction f)

    {

      for (; first != last; ++first) {

        f(*first);

      }

      return f;

    }

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

Issues with the code and Possible solution approaches below:

1. A Lambda expression is not std::function, it can only be stored in an object of type std::function.

Approach: (With your existing signature):

std::function<void()> f = () -> void { // do something };

for_each(vec.begin(), vec.end(), f);

2. Better solution would be to use a Callable as a parameter (Go to end of the answer).

Does all you want and is also consistent with other standard algorithms: en.cppreference.com/algorithms/for_each:

    template<class InputIt, class UnaryFunction>

    constexpr UnaryFunction for_each(InputIt first, InputIt last, UnaryFunction f)

    {

      for (; first != last; ++first) {

        f(*first);

      }

      return f;

    }

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

edited Dec 29 '18 at 16:39

answered Dec 29 '18 at 16:07

Optimaton

3815

answered Dec 29 '18 at 16:07

Optimaton

3815

answered Dec 29 '18 at 16:07

Optimaton

3815

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

add a comment |

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

Using a forwarding reference isn't going to prevent you from being "able to reuse the function object" unless you forward the reference. And using an lvalue reference instead prevents you from calling your function with a lambda as an argument for_each(x, y, (...){...});, which is not a good thing...

– HolyBlackCat
Dec 29 '18 at 16:15

@HolyBlackCat Incorporated the suggested changes. Must have sneaked in as I had to type it in due to some issues while copying. :p

– Optimaton
Dec 29 '18 at 16:21

add a comment |

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