Re: What to do about default function arguments

[Martin Nowak]

On Thu, 26 Apr 2012 06:10:14 +0200, Walter Bright  
<newshou...@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; }
}