• "Dan" <dbdavidson yahoo.com> Nov 06 2012
• =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> Nov 06 2012
• "Dan" <dbdavidson yahoo.com> Nov 06 2012

"Dan" <dbdavidson yahoo.com> writes:

Why does g(cs) compile but not G(cs)?

Thanks,
Dan

struct S{}

void g(ref S i) {}
void g(const ref S i) {}
void G(T)(ref T i) {}
void G(T)(const ref T i) {}

void main() {
   S s;
   const(S) cs;
   g(s);
   g(cs);
   G(s);
   // Error: template p.G matches more than one template 
declaration, /p.d(5):G(T) and /p.d(6):G(T)
   // G(cs);
}

=?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:

On 11/06/2012 09:18 AM, Dan wrote:
 Why does g(cs) compile but not G(cs)?

 Thanks,
 Dan

 struct S{}

 void g(ref S i) {}
 void g(const ref S i) {}
 void G(T)(ref T i) {}
 void G(T)(const ref T i) {}

 void main() {
 S s;
 const(S) cs;
 g(s);
 g(cs);
 G(s);
 // Error: template p.G matches more than one template declaration,
 /p.d(5):G(T) and /p.d(6):G(T)
 // G(cs);
 }


Because in the case of cs, T is 'const(S)', matching the following two 
instantiations:

1) ref const(S) i
2) const ref const(S) i

The second line has a redundant const. So, they end up having the same 
signature.

Ali

"Dan" <dbdavidson yahoo.com> writes:

On Tuesday, 6 November 2012 at 18:34:04 UTC, Ali Çehreli wrote:

 Because in the case of cs, T is 'const(S)', matching the 
 following two instantiations:

 1) ref const(S) i
 2) const ref const(S) i

 The second line has a redundant const. So, they end up having 
 the same signature.

 Ali


Oh - that clears it up, thanks. G(T)(ref T i) is really 
potentially many functions whereas g(ref S i) is only one and no 
conflict. What is the standard approach in the face of templates 
to differentiate between const(T) and T?

If templates do you always do something like:

void G(T)(ref T i) if(isMutable!T) { }
void G(T)(const ref T i) { }

Thanks,
Dan