Parameters declared as the alias of a template won't accept the arguments of the same type.

[Tejas]

On Monday, 2 May 2022 at 22:01:51 UTC, ag0aep6g wrote:
> On 02.05.22 22:47, Stanislav Blinov wrote:
>> On Monday, 2 May 2022 at 20:16:04 UTC, ag0aep6g wrote:
>>> On 02.05.22 21:17, Stanislav Blinov wrote:
>>>> On Monday, 2 May 2022 at 16:29:05 UTC, Loara wrote:
>>> [...]
>>>>> ```d
>>>>>     template MyAlias(T){
>>>>>       alias MyAlias = int;
>>>>>     }
>>>>>
>>>>>     T simp(T)(MyAlias!T val){
>>>>>       return T.init;
>>>>>     }
>>>>>
>>>>>     int main(){
>>>>>       simp(3);//Impossible to deduce T
> [...]
>> That's not my answer. And it is nonsense, because my answer is 
>> - what T is irrelevant, MyAlias maps *all* types to int. 
>> Therefore `simp` takes an int, which is what the caller is 
>> passing.
>
> And what does it return?

Perhaps one way could be to assume that the parameter passed is 
of type `aliasTemplate!typeOfVar` and see if it helps make sense 
of the invocation? This would only be done if the parameter's 
type doesn't directly match the required type nor it is 
implicitly convertible to it(which can be further constrained 
by... constraints...)
For example, here:
```d
template MyAlias(T){
       alias MyAlias = int;
     }
     // Please note that I changed code, since this is the only 
way to get the type with which the template parameter was 
initialized
     U simp(T:MyAlias!U, U)(auto ref T val){
       return U.init;
     }

     int main(){
       simp(3); // val is inferred to be of type MyAlias!int
       simp(”4”); // val is of type MyAlias!string
       simp!string([4,0]); val is of type MyAlias!(int[])
       MyAlias!float d =45;
       simp(d); // val is inferred to be of type 
MyAlias!(MyAlias!float) // hopefully though the second MyAlias 
doesn't have to exist and it can just assume the type to be int
     }
```