Why is `const int& k = i; ++i; ` possible?












6















I am supposed to determine whether this function is syntactically correct:



int f3(int i, int j) { const int& k=i; ++i; return k; }



I have tested it out and it compiles with my main function.



I do not understand why this is so.



Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



Any help is greatly appreciated






c++ syntax reference const







share|improve this question



























    6















    I am supposed to determine whether this function is syntactically correct:



    int f3(int i, int j) { const int& k=i; ++i; return k; }



    I have tested it out and it compiles with my main function.



    I do not understand why this is so.



    Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



    Any help is greatly appreciated






    c++ syntax reference const







    share|improve this question

























      6












      6








      6








      I am supposed to determine whether this function is syntactically correct:



      int f3(int i, int j) { const int& k=i; ++i; return k; }



      I have tested it out and it compiles with my main function.



      I do not understand why this is so.



      Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



      Any help is greatly appreciated






      c++ syntax reference const







      share|improve this question














      I am supposed to determine whether this function is syntactically correct:



      int f3(int i, int j) { const int& k=i; ++i; return k; }



      I have tested it out and it compiles with my main function.



      I do not understand why this is so.



      Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



      Any help is greatly appreciated






      c++ syntax reference const




      c++ syntax reference const






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 7 hours ago









      user9078057user9078057

      1118




      1118
























          1 Answer
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          14















          the increment ++iwill result in ++k which is not possible given that it was set const




          That's a misunderstanding.



          You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



          Here's a far-fetched analogy.



          You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
          outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






          share|improve this answer





















          • 5





            Not so far-fetched. That's a good analogy.

            – user4581301
            6 hours ago











          • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

            – David Schwartz
            6 hours ago











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          active

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          14















          the increment ++iwill result in ++k which is not possible given that it was set const




          That's a misunderstanding.



          You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



          Here's a far-fetched analogy.



          You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
          outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






          share|improve this answer





















          • 5





            Not so far-fetched. That's a good analogy.

            – user4581301
            6 hours ago











          • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

            – David Schwartz
            6 hours ago
















          14















          the increment ++iwill result in ++k which is not possible given that it was set const




          That's a misunderstanding.



          You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



          Here's a far-fetched analogy.



          You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
          outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






          share|improve this answer





















          • 5





            Not so far-fetched. That's a good analogy.

            – user4581301
            6 hours ago











          • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

            – David Schwartz
            6 hours ago














          14












          14








          14








          the increment ++iwill result in ++k which is not possible given that it was set const




          That's a misunderstanding.



          You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



          Here's a far-fetched analogy.



          You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
          outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






          share|improve this answer
















          the increment ++iwill result in ++k which is not possible given that it was set const




          That's a misunderstanding.



          You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



          Here's a far-fetched analogy.



          You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
          outside. You have const access or view access to the store while the employee has non-const access or change access to the store.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          R SahuR Sahu

          169k1294193




          169k1294193








          • 5





            Not so far-fetched. That's a good analogy.

            – user4581301
            6 hours ago











          • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

            – David Schwartz
            6 hours ago














          • 5





            Not so far-fetched. That's a good analogy.

            – user4581301
            6 hours ago











          • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

            – David Schwartz
            6 hours ago








          5




          5





          Not so far-fetched. That's a good analogy.

          – user4581301
          6 hours ago





          Not so far-fetched. That's a good analogy.

          – user4581301
          6 hours ago













          Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

          – David Schwartz
          6 hours ago





          Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

          – David Schwartz
          6 hours ago




















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