Re: How do you call an eponymous template that has a secondary template arg?
On Monday, 12 March 2018 at 04:15:23 UTC, Simen Kjærås wrote:
Yeah, that's a little hole in the grammar, but there are ways:
// Declare an alias:
alias aliasOfInt = aliasOf!int;
// And use that:
assert(!aliasOfInt!string);
Or use std.meta.Instantiate:
assert(!Instantiate!(aliasOf!int, string));
--
Simen
Noice! Did not know about instantiate.
Btw: I just tried this and it worked, to my surprise!:
template aliasOf(T) {
auto aliasOf(U)(U) { return is(U == T); }
enum aliasOf(alias a) = aliasOf!int(a);
}
void main() {
writeln(aliasOf!int(3)); // works
writeln(allSatisfy!(aliasOf!int, 3)); // works
}
Re: How do you call an eponymous template that has a secondary template arg?
On Sunday, 11 March 2018 at 12:05:56 UTC, aliak wrote: * aliasOf!int!"string" // multiple ! arguments are not allowed * (aliasOf!int)!"string" // error c-style cast * aliasOf!int.aliasOf!"string" // template isAliasOf(alias a) does not have property 'isAliasOf Yeah, that's a little hole in the grammar, but there are ways: // Declare an alias: alias aliasOfInt = aliasOf!int; // And use that: assert(!aliasOfInt!string); Or use std.meta.Instantiate: assert(!Instantiate!(aliasOf!int, string)); -- Simen
Re: How do you call an eponymous template that has a secondary template arg?
On Sunday, 11 March 2018 at 13:44:38 UTC, Basile B. wrote:
The first version works here:
```
template aliasOf(T) {
enum aliasOf(alias a) = is(typeof(a) == T);
}
string s;
pragma(msg, allSatisfy!(aliasOf!string, s, "string"));
```
I can see that my description was a little confusing, sorry about
that, I meant to ask how would you call that without using the
allSatisfy meta template. If I were to call it as a stand alone,
ie:
writeln(aliasOf!stringtemplate?>);
I hope that makes it clearer.
Now on the fact that what is done is correct is another story.
If the literal passed is supposed to be a type then it's
clearly wrong.
You'd have to mix it:
```
template aliasOf(T)
{
template aliasOf(alias a)
{
mixin("alias A = " ~ a ~ ";");
enum aliasOf = is(A == T);
}
}
alias myString1 = string;
alias myString2 = string;
pragma(msg, allSatisfy!(aliasOf!string, "myString1",
"myString2"));
```
Aye, I see what you mean, but it is supposed to be a literal of a
specific type. I.e. 3, "some string", SomeType(), etc.
So aliasOf!int.aliasOf!3 // true
Also, if I define it like this:
template aliasOf(T) {
auto aliasOf(U)(U) { return is(U == T); }
}
Then at least I can call it like:
writeln(aliasOf!int(3)); // prints true
But then I can't do:
allSatisfy!(aliasOf!int, 3)
I guess because it's a function now and not a template anymore so
can't be used by allSatisfy.
Re: How do you call an eponymous template that has a secondary template arg?
On Sunday, 11 March 2018 at 12:05:56 UTC, aliak wrote:
Eg:
template aliasOf(T) {
enum aliasOf(alias a) = is(typeof(a) == T);
}
The use case for this is for std.meta.allSatisfy for variadic
args, i.e.
template T(values...) if (allSatisfy!(aliasOf!string, values) {
... }
But how do you call that template otherwise?
I've tries:
* aliasOf!int!"string" // multiple ! arguments are not allowed
* (aliasOf!int)!"string" // error c-style cast
* aliasOf!int.aliasOf!"string" // template isAliasOf(alias a)
does not have property 'isAliasOf
I can work around this by:
template typeOf(T) {
enum isAliasedBy(alias a) = is(typeof(a) == T);
}
and then do:
template T(values...) if
(allSatisfy!(typeOf!string.isAliasedBy, values) { ... }
But I like the readability of the former better if there's a
way to achieve it?
Cheers
- Ali
The first version works here:
```
template aliasOf(T) {
enum aliasOf(alias a) = is(typeof(a) == T);
}
string s;
pragma(msg, allSatisfy!(aliasOf!string, s, "string"));
```
Now on the fact that what is done is correct is another story.
If the literal passed is supposed to be a type then it's clearly
wrong.
You'd have to mix it:
```
template aliasOf(T)
{
template aliasOf(alias a)
{
mixin("alias A = " ~ a ~ ";");
enum aliasOf = is(A == T);
}
}
alias myString1 = string;
alias myString2 = string;
pragma(msg, allSatisfy!(aliasOf!string, "myString1",
"myString2"));
```
