C++ lambda syntax












10















I have a function that searches a vector of iterators and returns the iterator if its names matches a string passed as an argument.



koalaGraph::PVertex lookUpByName(std::string Name, std::vector<koalaGraph::PVertex>& Vertices) {

for (size_t i = 0; i < Vertices.size(); i++) {

if(Vertices[i]->info.name == Name)
return Vertices[i];
}
}


My question is how can I implement this as a lambda, to use it in connection with std::find_if?



I'm trying this:



std::vector<koalaGraph::PVertex> V;
std::string Name;
std::find_if(V.begin(), V.end(), [&Name]() {return Name == V->info.name;})


But it says that V




an enclosing-function local variable cannot be referenced in a lambda body unless it is in the capture list.











share|improve this question









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  • 1





    No return if nothing found?

    – Deduplicator
    yesterday
















10















I have a function that searches a vector of iterators and returns the iterator if its names matches a string passed as an argument.



koalaGraph::PVertex lookUpByName(std::string Name, std::vector<koalaGraph::PVertex>& Vertices) {

for (size_t i = 0; i < Vertices.size(); i++) {

if(Vertices[i]->info.name == Name)
return Vertices[i];
}
}


My question is how can I implement this as a lambda, to use it in connection with std::find_if?



I'm trying this:



std::vector<koalaGraph::PVertex> V;
std::string Name;
std::find_if(V.begin(), V.end(), [&Name]() {return Name == V->info.name;})


But it says that V




an enclosing-function local variable cannot be referenced in a lambda body unless it is in the capture list.











share|improve this question









New contributor




black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1





    No return if nothing found?

    – Deduplicator
    yesterday














10












10








10


1






I have a function that searches a vector of iterators and returns the iterator if its names matches a string passed as an argument.



koalaGraph::PVertex lookUpByName(std::string Name, std::vector<koalaGraph::PVertex>& Vertices) {

for (size_t i = 0; i < Vertices.size(); i++) {

if(Vertices[i]->info.name == Name)
return Vertices[i];
}
}


My question is how can I implement this as a lambda, to use it in connection with std::find_if?



I'm trying this:



std::vector<koalaGraph::PVertex> V;
std::string Name;
std::find_if(V.begin(), V.end(), [&Name]() {return Name == V->info.name;})


But it says that V




an enclosing-function local variable cannot be referenced in a lambda body unless it is in the capture list.











share|improve this question









New contributor




black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have a function that searches a vector of iterators and returns the iterator if its names matches a string passed as an argument.



koalaGraph::PVertex lookUpByName(std::string Name, std::vector<koalaGraph::PVertex>& Vertices) {

for (size_t i = 0; i < Vertices.size(); i++) {

if(Vertices[i]->info.name == Name)
return Vertices[i];
}
}


My question is how can I implement this as a lambda, to use it in connection with std::find_if?



I'm trying this:



std::vector<koalaGraph::PVertex> V;
std::string Name;
std::find_if(V.begin(), V.end(), [&Name]() {return Name == V->info.name;})


But it says that V




an enclosing-function local variable cannot be referenced in a lambda body unless it is in the capture list.








c++ lambda






share|improve this question









New contributor




black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited yesterday









jvb

1133




1133






New contributor




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asked yesterday









black sheepblack sheep

603




603




New contributor




black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






black sheep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1





    No return if nothing found?

    – Deduplicator
    yesterday














  • 1





    No return if nothing found?

    – Deduplicator
    yesterday








1




1





No return if nothing found?

– Deduplicator
yesterday





No return if nothing found?

– Deduplicator
yesterday












5 Answers
5






active

oldest

votes


















18














find_if is going to pass the elements of the vector into your lambda. That means you need



std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})


so that the V in the lambda body is the element of the vector, not the vector itself.





Ideally you'd give it a different name than V so you keep the vector and local variables separate like



std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})


So now it is clear you are working on a element of the vector, instead of the vector itself.






share|improve this answer





















  • 4





    It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

    – Handy999
    yesterday











  • @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

    – Rakete1111
    yesterday











  • @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

    – NathanOliver
    yesterday



















7














First, V->info.name is ill formed, inside or outside of the lambda.



The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.



auto found = std::find_if(
V.begin(), V.end(),
[&Name](koalaGraph::PVertex const& item_to_check) {
return Name == item_to_check->info.name;
}
);


The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()



If you use C++14 or better, you can even use generic lambdas:



auto found = std::find_if(
V.begin(), V.end(),
[&Name](auto const& item_to_check) {
return Name == item_to_check->info.name;
}
);





share|improve this answer


























  • V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

    – Silvio Mayolo
    yesterday






  • 2





    It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

    – Guillaume Racicot
    yesterday











  • Not all compile errors are syntax errors. See syntax vs. semantics.

    – Silvio Mayolo
    yesterday






  • 1





    @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

    – Guillaume Racicot
    yesterday





















4














std::find_if's predicate will receive a reference to each element of the range in turn. You need:



std::find_if(
V.begin(), V.end(),
[&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
);





share|improve this answer































    2














    Get V as parameter to the lambda.



    std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)






    share|improve this answer































      1














      Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.



      struct PVertex
      {
      std::string name;
      };

      int main()
      {
      std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

      std::string Name = "cat";

      auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

      std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
      }


      Result is:



      found at: V[2]


      Example: https://rextester.com/IYNA58046






      share|improve this answer























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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        18














        find_if is going to pass the elements of the vector into your lambda. That means you need



        std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})


        so that the V in the lambda body is the element of the vector, not the vector itself.





        Ideally you'd give it a different name than V so you keep the vector and local variables separate like



        std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})


        So now it is clear you are working on a element of the vector, instead of the vector itself.






        share|improve this answer





















        • 4





          It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

          – Handy999
          yesterday











        • @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

          – Rakete1111
          yesterday











        • @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

          – NathanOliver
          yesterday
















        18














        find_if is going to pass the elements of the vector into your lambda. That means you need



        std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})


        so that the V in the lambda body is the element of the vector, not the vector itself.





        Ideally you'd give it a different name than V so you keep the vector and local variables separate like



        std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})


        So now it is clear you are working on a element of the vector, instead of the vector itself.






        share|improve this answer





















        • 4





          It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

          – Handy999
          yesterday











        • @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

          – Rakete1111
          yesterday











        • @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

          – NathanOliver
          yesterday














        18












        18








        18







        find_if is going to pass the elements of the vector into your lambda. That means you need



        std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})


        so that the V in the lambda body is the element of the vector, not the vector itself.





        Ideally you'd give it a different name than V so you keep the vector and local variables separate like



        std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})


        So now it is clear you are working on a element of the vector, instead of the vector itself.






        share|improve this answer















        find_if is going to pass the elements of the vector into your lambda. That means you need



        std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})


        so that the V in the lambda body is the element of the vector, not the vector itself.





        Ideally you'd give it a different name than V so you keep the vector and local variables separate like



        std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})


        So now it is clear you are working on a element of the vector, instead of the vector itself.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered yesterday









        NathanOliverNathanOliver

        95.8k16135208




        95.8k16135208








        • 4





          It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

          – Handy999
          yesterday











        • @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

          – Rakete1111
          yesterday











        • @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

          – NathanOliver
          yesterday














        • 4





          It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

          – Handy999
          yesterday











        • @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

          – Rakete1111
          yesterday











        • @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

          – NathanOliver
          yesterday








        4




        4





        It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

        – Handy999
        yesterday





        It would be better not to use the same name V for two different things (the vector and the iterator). Thus, std::find_if(V.begin(), V.end(), [&Name](auto const& it) {return Name == it->info.name;}) would be better.

        – Handy999
        yesterday













        @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

        – Rakete1111
        yesterday





        @Handy999 it is also misleading though, since it's not an iterator. Maybe value, element or whatever else :)

        – Rakete1111
        yesterday













        @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

        – NathanOliver
        yesterday





        @Handy999 You're right about giving it a better name, I just wanted to keep it as V as it makes my last sentence very succinct.

        – NathanOliver
        yesterday













        7














        First, V->info.name is ill formed, inside or outside of the lambda.



        The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](koalaGraph::PVertex const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );


        The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()



        If you use C++14 or better, you can even use generic lambdas:



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](auto const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );





        share|improve this answer


























        • V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

          – Silvio Mayolo
          yesterday






        • 2





          It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

          – Guillaume Racicot
          yesterday











        • Not all compile errors are syntax errors. See syntax vs. semantics.

          – Silvio Mayolo
          yesterday






        • 1





          @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

          – Guillaume Racicot
          yesterday


















        7














        First, V->info.name is ill formed, inside or outside of the lambda.



        The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](koalaGraph::PVertex const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );


        The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()



        If you use C++14 or better, you can even use generic lambdas:



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](auto const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );





        share|improve this answer


























        • V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

          – Silvio Mayolo
          yesterday






        • 2





          It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

          – Guillaume Racicot
          yesterday











        • Not all compile errors are syntax errors. See syntax vs. semantics.

          – Silvio Mayolo
          yesterday






        • 1





          @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

          – Guillaume Racicot
          yesterday
















        7












        7








        7







        First, V->info.name is ill formed, inside or outside of the lambda.



        The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](koalaGraph::PVertex const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );


        The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()



        If you use C++14 or better, you can even use generic lambdas:



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](auto const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );





        share|improve this answer















        First, V->info.name is ill formed, inside or outside of the lambda.



        The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](koalaGraph::PVertex const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );


        The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()



        If you use C++14 or better, you can even use generic lambdas:



        auto found = std::find_if(
        V.begin(), V.end(),
        [&Name](auto const& item_to_check) {
        return Name == item_to_check->info.name;
        }
        );






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered yesterday









        Guillaume RacicotGuillaume Racicot

        15.4k53568




        15.4k53568













        • V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

          – Silvio Mayolo
          yesterday






        • 2





          It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

          – Guillaume Racicot
          yesterday











        • Not all compile errors are syntax errors. See syntax vs. semantics.

          – Silvio Mayolo
          yesterday






        • 1





          @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

          – Guillaume Racicot
          yesterday





















        • V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

          – Silvio Mayolo
          yesterday






        • 2





          It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

          – Guillaume Racicot
          yesterday











        • Not all compile errors are syntax errors. See syntax vs. semantics.

          – Silvio Mayolo
          yesterday






        • 1





          @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

          – Guillaume Racicot
          yesterday



















        V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

        – Silvio Mayolo
        yesterday





        V->info.name is valid syntax. It just won't typecheck since vectors don't overload ->.

        – Silvio Mayolo
        yesterday




        2




        2





        It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

        – Guillaume Racicot
        yesterday





        It just won't typecheck since vectors don't overload -> so... it won't compile? It's an invalid syntax to use -> on object of types that don't overload operator->

        – Guillaume Racicot
        yesterday













        Not all compile errors are syntax errors. See syntax vs. semantics.

        – Silvio Mayolo
        yesterday





        Not all compile errors are syntax errors. See syntax vs. semantics.

        – Silvio Mayolo
        yesterday




        1




        1





        @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

        – Guillaume Racicot
        yesterday







        @SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly

        – Guillaume Racicot
        yesterday













        4














        std::find_if's predicate will receive a reference to each element of the range in turn. You need:



        std::find_if(
        V.begin(), V.end(),
        [&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
        );





        share|improve this answer




























          4














          std::find_if's predicate will receive a reference to each element of the range in turn. You need:



          std::find_if(
          V.begin(), V.end(),
          [&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
          );





          share|improve this answer


























            4












            4








            4







            std::find_if's predicate will receive a reference to each element of the range in turn. You need:



            std::find_if(
            V.begin(), V.end(),
            [&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
            );





            share|improve this answer













            std::find_if's predicate will receive a reference to each element of the range in turn. You need:



            std::find_if(
            V.begin(), V.end(),
            [&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
            );






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered yesterday









            QuentinQuentin

            46.6k589146




            46.6k589146























                2














                Get V as parameter to the lambda.



                std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)






                share|improve this answer




























                  2














                  Get V as parameter to the lambda.



                  std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)






                  share|improve this answer


























                    2












                    2








                    2







                    Get V as parameter to the lambda.



                    std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)






                    share|improve this answer













                    Get V as parameter to the lambda.



                    std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered yesterday









                    Petar VelevPetar Velev

                    1,681619




                    1,681619























                        1














                        Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.



                        struct PVertex
                        {
                        std::string name;
                        };

                        int main()
                        {
                        std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

                        std::string Name = "cat";

                        auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

                        std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
                        }


                        Result is:



                        found at: V[2]


                        Example: https://rextester.com/IYNA58046






                        share|improve this answer




























                          1














                          Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.



                          struct PVertex
                          {
                          std::string name;
                          };

                          int main()
                          {
                          std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

                          std::string Name = "cat";

                          auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

                          std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
                          }


                          Result is:



                          found at: V[2]


                          Example: https://rextester.com/IYNA58046






                          share|improve this answer


























                            1












                            1








                            1







                            Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.



                            struct PVertex
                            {
                            std::string name;
                            };

                            int main()
                            {
                            std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

                            std::string Name = "cat";

                            auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

                            std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
                            }


                            Result is:



                            found at: V[2]


                            Example: https://rextester.com/IYNA58046






                            share|improve this answer













                            Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.



                            struct PVertex
                            {
                            std::string name;
                            };

                            int main()
                            {
                            std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

                            std::string Name = "cat";

                            auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

                            std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
                            }


                            Result is:



                            found at: V[2]


                            Example: https://rextester.com/IYNA58046







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered yesterday









                            Constantinos GlynosConstantinos Glynos

                            1,581820




                            1,581820






















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