argument of a function

To C++ experts,

 I have a code as follows :
--------------------------------------------------------------
class A{} ;
class B
{   public :
      void f(A const& a){ cout << "f" << endl ;}
      void g(A const&){ cout << "g" << endl ;}
};

int main()
{
  A a1 ;
  B b1 ;
  b1.f(a1) ;
  b1.g(a1) ;
  return 0 ;
}
---------------------------------------------
  It seems that both f(A const& a) and g(A const&) work. However, I am wondering is there any difference between them ? which one is better and which one should I use in general ?

  Thanks a lot !

meow.
 
LVL 1
meow00Asked:
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AxterCommented:
>>It seems that both f(A const& a) and g(A const&) work. However, I am wondering is there any difference between them ? which one is better and which one should I use in general ?

Yes.
Since function g doesn't have the argument name, the argument is not available for use within the function.

This type of declaration is commonly used to avoid compiler warnings or errors.
Example

int foo(int xyz)
{
  return 3;
}

Some compilers will give a warning for the above function.
*** Waring variable xyz declared but not used *****

To avoid such a warning, you could declare the function without the argument name.

int foo(int)
{
  return 3;
}

Now the compiler will not give a warning for unused variable.

Of course, if you don't need the variable, then you shouldn't have it in the function declaration.
However, this method is usefull when a function argument is no longer needed but you dont' wish to change all the code that is calling the function.

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Sys_ProgCommented:
Just to add, the above technique is termed as having Place-Holder Arguments

It is used when u know want to keep some dummy parameter to a function, so that u could use it later without affecting the client code using that function

HTH

Amit

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