TITLE: Method name scoping example

[ This example illustrates a very general concept missed by many
  new C++ programmers. -adc
]

PROBLEM: davidm@consilium.com (David S. Masterson)

 4      class alpha { 
 8      public: 
 9              ~alpha() {} 
14      private: 
15              alpha(int i = 0) : x(i) {} 


[	        void* copy() { return new alpha; }

  The above line was missing from the posting. I tried to reconstruct
  a line which would show the same problem. -adc ]


===1> 153: Error: Could not find a match for alpha::operator new(unsigned int)

16              void* operator new(size_t sz, void* tmp) 
17                      { return ::new char[sz]; } 
19      }; 


RESPONSE: gs4t@virginia.edu (Gnanasekaran Swaminathan), 22 Sep 92

This exact case is discussed in an example in p282 of ARM.


RESPONSE: davidm@consilium.com (David S. Masterson)

Ah, yes.  What I don't understand, though, is why the above 'new' definition
hides the standard global 'new' definition.  The function signature doesn't
seem the same to me.


RESPONSE: gs4t@virginia.edu (Gnanasekaran Swaminathan)

For the same reason as to why the following example doesn't work.

void f(int) {} // f(int) is in global scope

struct A {
	static void f(int b, int c) {} // hide ::f
	int a() {
		f(10);     // error as ::f(int) is hidden
		::f(10);   // ok 
		f(10, 20); // A::f(int, int)
	}
};

ARM p.15 says, "A name may be hidden by an explicit declaration
of that same name in an enclosed block or in a class."

Hence, in your case, by declaring alpha::operator new(size_t, void*)
you have hidden ::operator new(size_t)