Why the default arguments of overloaded functions are not considered by the c++ come by there as part of the parameter list?

2/3/2017 7:12:20 AM

Srikanth Srinivasan

4 Answers

New Answer


This is actually a good question! Lets consider what happens if this was possible (we'll overload the assignment operator): class C { public: C(int x){this->x=x;} C& operator=(const C& rhs=C(0)) //Note default argument {x = rhs.x; return *this;} private: int x; }; The usual way this works: C c1(1); C c2(2); c1 = c2; //Invokes c1.operator=(c2); c1.x is now 2 //Lets try to use default value now: c1 = ???; /*How do we write this statement to omit the rhs to allow us to call c1.operator=(); which will use the default argument?*/ In short, there is no legitimate syntax of the form c1 = ???; to allow this. If the compiler allows c1 = ; as syntax for this - then an error might pass as valid code! For example this c1 = ;c2; might be an innocent typo but will now be valid and have totally unexpected and wrong results.


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