Re: Function pointers/delegates default args were stealth removed?

[Timon Gehr]

On 08/28/2012 10:33 PM, Carl Sturtivant wrote:

On Monday, 27 August 2012 at 00:44:54 UTC, Walter Bright wrote:

On 8/26/2012 4:50 PM, Timon Gehr wrote:

On 08/27/2012 12:41 AM, Walter Bright wrote:

The trouble for function pointers, is that any default args would need
to be part of the type, not the declaration.

They could be made part of the variable declaration.

You mean part of the function pointer variable?

Consider what you do with a function pointer - you pass it to someone
else. That someone else gets it as a type, not a declaration. I.e. you
lose the default argument information, since that is not attached to
the type.

I think this is the right behavior too. Default arguments are IMHO just
a compact way to write some simply related overloaded functions, e.g. thus:

   int sum(int x, int y = 1 ) { return x + y; }

is just a compact way to write

   int sum(int x, int y) { return x + y; }
   int sum(int x) { return sum(x, 1); }
...

This interpretation is simply wrong.

import std.stdio, std.c.stdlib;

void* alloca20bytes(void* x = alloca(20)){ return x; }

// your suggested transformation:

/+void* alloca20bytes(void* x){ return x; }
void* alloca20bytes(){ return alloca20bytes(alloca(20)); }+/

// would break the caller:

void main(){
    auto x = (cast(int*)alloca20bytes())[0..5];
    x[] = 0;
    x[] += 2;
    writeln(x);
}