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Default Arguments are not Part of a Function’s Type

Default Arguments are not Part of a Function’s Type

Although the default arguments of a function must appear in its declaration, they are not considered part of its type. Thus, you cannot overload a function by using different default arguments:

           void f(int n = 6);  void f(int n = 0); //error, redefinition of f() above

The attempt to overload f() is illegal because the compiler cannot distinguish between the two versions of f().

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