Comment #0 by e.insafutdinov — 2009-12-25T01:02:49Z
void foo(Fn)(Fn fn)
{
fn();
}
void bar(int i = 22) {
writeln("bar ", i);
}
void bam(int i) {
writeln("bam ", i);
}
void baz() {
foo(&bar);
foo(&bam);
}
This code compiles while it should not. You cannot call bam() without any arguments. If you switch the instantiations and call foo(&bam) first - it does not compile.
pragma(msg, typeof(&bar).stringof);
pragma(msg, typeof(&bam).stringof);
outputs
void function(int i = 22)
void function(int i)
But it is not respected by the template.
Comment #1 by 2korden — 2009-12-25T08:32:32Z
This is a bit more tricky than it looks like.
The question here is, how many times should foo be instanciated?
DMD answers "just 1 time", and I agree with it, since bar and bam have same signature and mangling.
Having only one instantiation means that code for it will also be generated once.
When you call a function that has default argument without providing that argument, the call is actually *rewritten to include that argument*. I.e. bar(); is rewritten as bar(22);
Not lets see how foo!(bar) looks like after a rewrite:
void foo(Fn)(Fn fn)
{
fn(22);
}
Now there is no wonder why dmd behaves like this.
But this only happens if foo!(bar) is instanciated before foo!(bam), because when templates are instantiates on first use. When you instanciate foo!(bam) first, it doesn't get rewritten and therefore fails to compile.
I guess the only fix for this issue would be to create an implicit trampoline to invoke functions with default arguments, but only Walter can say for sure.
Comment #2 by e.insafutdinov — 2009-12-26T12:51:57Z
(In reply to comment #1)
> This is a bit more tricky than it looks like.
>
> The question here is, how many times should foo be instanciated?
> DMD answers "just 1 time", and I agree with it, since bar and bam have same
> signature and mangling.
> But this only happens if foo!(bar) is instanciated before foo!(bam), because
> when templates are instantiates on first use. When you instanciate foo!(bam)
> first, it doesn't get rewritten and therefore fails to compile.
>
> I guess the only fix for this issue would be to create an implicit trampoline
> to invoke functions with default arguments, but only Walter can say for sure.
Yeah, that was pretty much what I was thinking about. The issue cuts down to a question, should Foo be instantiated once or are the function types equal? For binary size and efficiency purposes I agree that there should be one instantiation. On the other hand for meta programming and code generation which is what I am doing I would like to have them as separate types, as I don't want to loose this information. Walter or Andrei would probably have to comment on this issue.
Comment #3 by smjg — 2010-01-03T08:53:39Z
I've never heard of default arguments being part of the function type's properties. I'd probably guessed that the type is simply void function(int) but default arguments are filled in on the caller side. In which case the functions are the same type, and so this isn't meant to work. Either way, it's certainly a bug that the code is accepted.
And it applies to D1 too, but the code needs changing a bit:
----------
import std.stdio;
void foo(Fn)(Fn fn) {
fn();
}
void bar(int i = 22) {
writefln("bar ", i);
}
void bam(int i) {
writefln("bam ", i);
}
void main() {
foo(&bar);
foo(&bam);
}
----------
Comment #4 by yebblies — 2011-07-03T05:22:51Z
*** Issue 4028 has been marked as a duplicate of this issue. ***
Comment #5 by yebblies — 2011-07-03T05:23:21Z
*** Issue 5456 has been marked as a duplicate of this issue. ***
(In reply to comment #1)
> Not lets see how foo!(bar) looks like after a rewrite:
>
> void foo(Fn)(Fn fn)
> {
> fn(22);
> }
Actually, before this rewrite comes substituting in the template argument.
void fooInstance(void function(int = 22) fn)
{
fn();
}
which then becomes
void fooInstance(void function(int) fn)
{
fn(22);
}
> Now there is no wonder why dmd behaves like this.
But it makes sense only if there are actually two template instances involved. The essence of the bug is that, after it has instantiated foo with argument void function(int = 22), the compiler matches any attempt to instantiate foo with a void function(int), with or without a default argument, to this instance.
(In reply to comment #2)
> Yeah, that was pretty much what I was thinking about. The issue
> cuts down to a question, should Foo be instantiated once or are the
> function types equal? For binary size and efficiency purposes I
> agree that there should be one instantiation. On the other hand
> for meta programming and code generation which is what I am doing I
> would like to have them as separate types, as I don't want to loose
> this information. Walter or Andrei would probably have to comment
> on this issue.
But this seems to me a rare use case. And you can do it with an alias template parameter, though this does create a separate instance for each function name:
----------
import std.stdio;
void foo(alias fn)() {
fn();
fn(42);
}
void bar(int i = 22) {
writefln("bar %d", i);
}
void bam(int i) {
writefln("bam %d", i);
}
void quux() {
writefln("quux");
}
void quux(int i) {
writefln("quux %d", i);
}
void main() {
foo!(bar)(); // works
//foo!(bam)(); // correctly triggers compiler error (1.068 Win32)
foo!(quux)(); // works with overloads as well
}
----------
I can see three possible ways to resolve this issue:
(a) Default arguments aren't part of the function type. Then only one template instance exists: foo!(void function(int)). The template instance is then illegal, since it tries to call a void function(int) with no arguments.
(b) Default arguments are part of the function type. Then there are two template instances:
foo(&bar) is foo!(void function(int = 22))
foo(&bam) is foo!(void function(int))
Then foo(&bar) is legal: foo(&bar)() calls bar(22)
and foo(&bam) is illegal
(c) Default arguments are part of the function type, but can be optimised away. Essentially, we detect while instantiating the template whether the template body makes use of a default argument. If so, create a separate instance for each default argument or absence thereof. Otherwise, create only one instance. Not sure whether this is desirable.
Comment #8 by yebblies — 2012-02-11T18:57:19Z
This has the same causes (and same solutions) as issue 3866.
Comment #9 by aldacron — 2012-06-17T22:02:35Z
Somehow I managed to change the settings on the wrong bug. Sorry for the noise.
Comment #10 by issues.dlang — 2012-07-02T08:34:37Z
*** Issue 8336 has been marked as a duplicate of this issue. ***