Re: How to stringify a template instantiation expression?
On Thursday, 1 March 2018 at 17:48:02 UTC, Simen Kjærås wrote:
On Thursday, 1 March 2018 at 16:46:30 UTC, Yuxuan Shui wrote:
[...]
string TemplateStringOf(T...)()
if (T.length == 1)
{
import std.traits : TemplateOf, TemplateArgsOf;
import std.meta : AliasSeq, staticMap;
import std.string : indexOf;
import std.conv : text;
static if (is(typeof({ alias a = TemplateOf!T; })))
{
alias Tmp = TemplateOf!T;
alias Args = TemplateArgsOf!T;
[...]
Ah, thanks. I was trying to match template instantiation using
is() (i.e. is(T == S!R, S, R...)). But this doesn't work if T is
a template.
Reading the code of TemplateOf, I realized I have to use template
constraints for that. Maybe we can add support for something like
is(alias T == ...)?
Re: How to stringify a template instantiation expression?
On Thursday, 1 March 2018 at 16:46:30 UTC, Yuxuan Shui wrote:
Did you actually try that? With dmd 2.079-rc1, this:
template A(T...) {}
struct B {}
struct D {}
struct E {}
template C(T...) {}
pragma(msg, (A!(B, C!(D, E))).stringof);
Prints:
A!(B, __T1CTS2ax1DTSQh1EZ)
When compiled
Yep, though not all possible combinations of course:
struct X(T...) {}
void main() {
writeln(X!(int, X!(X!(int), float), char).stringof);
}
Seems like if you throw in template templates then things get a
little more ... complicated.
Re: How to stringify a template instantiation expression?
On Thursday, 1 March 2018 at 16:46:30 UTC, Yuxuan Shui wrote:
On Wednesday, 28 February 2018 at 15:49:25 UTC, aliak wrote:
On Wednesday, 28 February 2018 at 15:09:56 UTC, Yuxuan Shui
wrote:
For a template instantiation expression like A!(B, C!(D, E)),
I want to get a string "A!(B, C!(D, E))", better if A, B, C,
D, E is replaced by fully qualified name.
Is this possible?
A!(B, C!(D, E)).stringof I guess. Will print the former.
There's a Learn forum as well btw :)
Cheers
Did you actually try that? With dmd 2.079-rc1, this:
template A(T...) {}
struct B {}
struct D {}
struct E {}
template C(T...) {}
pragma(msg, (A!(B, C!(D, E))).stringof);
Prints:
A!(B, __T1CTS2ax1DTSQh1EZ)
When compiled
string TemplateStringOf(T...)()
if (T.length == 1)
{
import std.traits : TemplateOf, TemplateArgsOf;
import std.meta : AliasSeq, staticMap;
import std.string : indexOf;
import std.conv : text;
static if (is(typeof({ alias a = TemplateOf!T; })))
{
alias Tmp = TemplateOf!T;
alias Args = TemplateArgsOf!T;
enum tmpFullName = Tmp.stringof;
enum tmpName = tmpFullName[0..tmpFullName.indexOf('(')];
alias AddCommas(T...) = AliasSeq!(T, ", ");
alias ArgNames = staticMap!(.TemplateStringOf, Args);
alias SeparatedArgNames = staticMap!(AddCommas,
ArgNames)[0..$-1];
return text(tmpName, "!(", SeparatedArgNames, ")");
}
else
{
return T[0].stringof;
}
}
unittest {
template A(T...) {}
struct B {}
struct D {}
struct E {}
template C(T...) {}
assert(TemplateStringOf!(A!(B, C!(D, E))) == "A!(B, C!(D,
E))");
}
unittest {
template A(T) { template A(T) {} }
assert(TemplateStringOf!(A!int) == "A!(int)");
}
Probably still some corner cases I haven't thought of, but it
seems to cover what you're asking for.
One thing I didn't bother with is disambiguation - if B exists in
modules foo and bar, the above will not specify which foo it's
referring to. This is left as an exercise for the reader.
--
Simen
Re: How to stringify a template instantiation expression?
On Thursday, 1 March 2018 at 16:46:30 UTC, Yuxuan Shui wrote:
On Wednesday, 28 February 2018 at 15:49:25 UTC, aliak wrote:
On Wednesday, 28 February 2018 at 15:09:56 UTC, Yuxuan Shui
wrote:
For a template instantiation expression like A!(B, C!(D, E)),
I want to get a string "A!(B, C!(D, E))", better if A, B, C,
D, E is replaced by fully qualified name.
Is this possible?
A!(B, C!(D, E)).stringof I guess. Will print the former.
There's a Learn forum as well btw :)
Cheers
Did you actually try that? With dmd 2.079-rc1, this:
template A(T...) {}
struct B {}
struct D {}
struct E {}
template C(T...) {}
pragma(msg, (A!(B, C!(D, E))).stringof);
Prints:
A!(B, __T1CTS2ax1DTSQh1EZ)
When compiled
Even worse, if the template instantiation yields another
template, e.g:
template A(T) { template A(T) {} }
A!int.stringof returns "A(T)", which is not useful at all.
Re: How to stringify a template instantiation expression?
On Wednesday, 28 February 2018 at 15:49:25 UTC, aliak wrote:
On Wednesday, 28 February 2018 at 15:09:56 UTC, Yuxuan Shui
wrote:
For a template instantiation expression like A!(B, C!(D, E)),
I want to get a string "A!(B, C!(D, E))", better if A, B, C,
D, E is replaced by fully qualified name.
Is this possible?
A!(B, C!(D, E)).stringof I guess. Will print the former.
There's a Learn forum as well btw :)
Cheers
Did you actually try that? With dmd 2.079-rc1, this:
template A(T...) {}
struct B {}
struct D {}
struct E {}
template C(T...) {}
pragma(msg, (A!(B, C!(D, E))).stringof);
Prints:
A!(B, __T1CTS2ax1DTSQh1EZ)
When compiled
Re: How to stringify a template instantiation expression?
On Wednesday, 28 February 2018 at 15:09:56 UTC, Yuxuan Shui wrote: For a template instantiation expression like A!(B, C!(D, E)), I want to get a string "A!(B, C!(D, E))", better if A, B, C, D, E is replaced by fully qualified name. Is this possible? A!(B, C!(D, E)).stringof I guess. Will print the former. There's a Learn forum as well btw :) Cheers
