TITLE: Template arguments and const


PROBLEM: stefans@bauv106.bauv.unibw-muenchen.de (Stefan Schwarz)

The following peace of code causes trouble when compiling:

template<class T> class A
{
private:
	T _a;
public:
	A(void) {}
	void set (const T a) { _a = a;}
};

main()
{
	A<int*> a;
	const int *x = new int(1);
	a.set(x);
}

error: bad argument  1 type for  A <int *>::set(): const int * 
( const int * expectd)

The problem doesn't occur with a class A<int>.



PROBLEM: vinoski@apollo.hp.com (Stephen Vinoski), 28 Sep 1992
          Hewlett-Packard


Parameter "a" to the set member function of the class template is of
type "const T".  Mentally making the substitution of "int *" for "T",
one might come up with:

	void set(const int *a) {_a = a;}

However, one would be wrong. :-) The template instantiation actually
results in:

	void set(int *const a) {_a = a;}

Apply const to the type T.  In this case, T is "int *", so we end up
with "a" being a "const pointer to int".

This is completely consistent with and similar to the treatment of
typedefs in the language.  For example:

	typedef int *IntPtr;
	const IntPtr p;

declares "p" to be a "const pointer to int", not a "pointer to const
int".

It might be easier to remind yourself of this behavior if you write
the declaration for the set member function like this:

	void set(T const a) {_a = a;}