Conditional ? : operator with class constructor











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could someone explain me why c and c1 are constructed different way.
I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance



#include <vector>

class foo {
public:
foo(const std::vector<int>& var) :var{ var } {};
const std::vector<int> & var;
};

std::vector<int> f(){
std::vector<int> x{ 1,2,3,4,5 };
return x;
};

int main(){
std::vector<int> x1{ 1,2,3,4,5 ,7 };
std::vector<int> x2{ 1,2,3,4,5 ,6 };
foo c{ true ? x2 : x1 }; //c.var has expected values
foo c1{ true ? x2 : f() }; //c.var empty
foo c2{ false ? x2 : f() }; //c.var empty
foo c3{ x2 }; //c.var has expected values
}









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    up vote
    10
    down vote

    favorite
    1












    could someone explain me why c and c1 are constructed different way.
    I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
    I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance



    #include <vector>

    class foo {
    public:
    foo(const std::vector<int>& var) :var{ var } {};
    const std::vector<int> & var;
    };

    std::vector<int> f(){
    std::vector<int> x{ 1,2,3,4,5 };
    return x;
    };

    int main(){
    std::vector<int> x1{ 1,2,3,4,5 ,7 };
    std::vector<int> x2{ 1,2,3,4,5 ,6 };
    foo c{ true ? x2 : x1 }; //c.var has expected values
    foo c1{ true ? x2 : f() }; //c.var empty
    foo c2{ false ? x2 : f() }; //c.var empty
    foo c3{ x2 }; //c.var has expected values
    }









    share|improve this question


























      up vote
      10
      down vote

      favorite
      1









      up vote
      10
      down vote

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      1





      could someone explain me why c and c1 are constructed different way.
      I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
      I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance



      #include <vector>

      class foo {
      public:
      foo(const std::vector<int>& var) :var{ var } {};
      const std::vector<int> & var;
      };

      std::vector<int> f(){
      std::vector<int> x{ 1,2,3,4,5 };
      return x;
      };

      int main(){
      std::vector<int> x1{ 1,2,3,4,5 ,7 };
      std::vector<int> x2{ 1,2,3,4,5 ,6 };
      foo c{ true ? x2 : x1 }; //c.var has expected values
      foo c1{ true ? x2 : f() }; //c.var empty
      foo c2{ false ? x2 : f() }; //c.var empty
      foo c3{ x2 }; //c.var has expected values
      }









      share|improve this question















      could someone explain me why c and c1 are constructed different way.
      I understand that I have reference to copy created by '?' operator, which is destroyed after construction, but why in first case it behave other way.
      I've tested if its optimization, but even with conditions read from console, I have same result. Thanks in advance



      #include <vector>

      class foo {
      public:
      foo(const std::vector<int>& var) :var{ var } {};
      const std::vector<int> & var;
      };

      std::vector<int> f(){
      std::vector<int> x{ 1,2,3,4,5 };
      return x;
      };

      int main(){
      std::vector<int> x1{ 1,2,3,4,5 ,7 };
      std::vector<int> x2{ 1,2,3,4,5 ,6 };
      foo c{ true ? x2 : x1 }; //c.var has expected values
      foo c1{ true ? x2 : f() }; //c.var empty
      foo c2{ false ? x2 : f() }; //c.var empty
      foo c3{ x2 }; //c.var has expected values
      }






      c++ c++14 conditional-operator






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      edited yesterday









      Deduplicator

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      geniculata

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          The type of a conditional expression is the common type of the two branches. For true ? x2 : x1, the common type is std::vector<int>&. For true ? x2 : f(), the common type is std::vector<int>.



          Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.






          share|improve this answer



















          • 4




            ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
            – Deduplicator
            yesterday






          • 5




            It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
            – Nicol Bolas
            yesterday










          • Oh C++. Oh you.
            – Lightness Races in Orbit
            yesterday











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          up vote
          18
          down vote













          The type of a conditional expression is the common type of the two branches. For true ? x2 : x1, the common type is std::vector<int>&. For true ? x2 : f(), the common type is std::vector<int>.



          Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.






          share|improve this answer



















          • 4




            ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
            – Deduplicator
            yesterday






          • 5




            It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
            – Nicol Bolas
            yesterday










          • Oh C++. Oh you.
            – Lightness Races in Orbit
            yesterday















          up vote
          18
          down vote













          The type of a conditional expression is the common type of the two branches. For true ? x2 : x1, the common type is std::vector<int>&. For true ? x2 : f(), the common type is std::vector<int>.



          Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.






          share|improve this answer



















          • 4




            ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
            – Deduplicator
            yesterday






          • 5




            It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
            – Nicol Bolas
            yesterday










          • Oh C++. Oh you.
            – Lightness Races in Orbit
            yesterday













          up vote
          18
          down vote










          up vote
          18
          down vote









          The type of a conditional expression is the common type of the two branches. For true ? x2 : x1, the common type is std::vector<int>&. For true ? x2 : f(), the common type is std::vector<int>.



          Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.






          share|improve this answer














          The type of a conditional expression is the common type of the two branches. For true ? x2 : x1, the common type is std::vector<int>&. For true ? x2 : f(), the common type is std::vector<int>.



          Therefore you are storing a dangling reference in c1. Any access to c1.var is undefined behavior.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited yesterday

























          answered yesterday









          Vittorio Romeo

          55.2k17146286




          55.2k17146286








          • 4




            ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
            – Deduplicator
            yesterday






          • 5




            It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
            – Nicol Bolas
            yesterday










          • Oh C++. Oh you.
            – Lightness Races in Orbit
            yesterday














          • 4




            ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
            – Deduplicator
            yesterday






          • 5




            It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
            – Nicol Bolas
            yesterday










          • Oh C++. Oh you.
            – Lightness Races in Orbit
            yesterday








          4




          4




          ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
          – Deduplicator
          yesterday




          ...resulting in UB, if it is accessed after the declaration. That crucial part was omitted from OP's example though.
          – Deduplicator
          yesterday




          5




          5




          It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
          – Nicol Bolas
          yesterday




          It'd be good to explain why these are the common types (ie: because the second case involves an lvalue and prvalue, while the first case is two lvalues).
          – Nicol Bolas
          yesterday












          Oh C++. Oh you.
          – Lightness Races in Orbit
          yesterday




          Oh C++. Oh you.
          – Lightness Races in Orbit
          yesterday


















           

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