What to do about default function arguments

[Martin Nowak]

On Thu, 26 Apr 2012 06:10:14 +0200, Walter Bright  
<newshound2 at digitalmars.com> wrote:

> On 4/25/2012 8:44 PM, Walter Bright wrote:
>> The problem centers around name mangling. If two types mangle the same,  
>> then
>> they are the same type. But default arguments are not part of the  
>> mangled
>> string. Hence the schizophrenic behavior.
>
> One might suggest mangling the default argument into the type. But  
> default arguments need not be compile time constants - they are  
> evaluated at runtime! Hence the unattractive specter of trying to mangle  
> a runtime expression.

import std.stdio;

int readVal()
{
     int val;
     stdin.readf("%s", &val);
     return val;
}

void main()
{
     auto dg = (int a=readVal()) => a;
     writeln(dg());
}

----

Stuffing the value into the type is not going to work out when taking the  
address.
I think it would be interesting to transform them to values of a type that
preserves the behavior. This would work for polymorphic lambdas as values  
too.

----
auto dg = (int a=readVal()) => a;

static struct Lamba
{
   int opCall() { return fbody(readVal()); }
   int opCall(int a) { return fbody(a); }
   int function(int) opAddrOf() { return &fbody; }
   static int fbody(int a) { return a; }
}

----
auto dg = a => 2 * a;

struct Lambda
{
   auto opCall(Ta)(auto ref Ta a) { return fbody(a); }
   @disable opAddrOf();
   /*static*/ auto fbody(Ta)(Ta a) { return 2 * a; }
}