Template constraints

[Tobias Pankrath via Digitalmars-d]

On Saturday, 14 February 2015 at 17:00:33 UTC, Andrei 
Alexandrescu wrote:
> There's been recurring discussion about failing constraints not 
> generating nice error messages.
>
> void fun(T)(T x) if (complicated_condition) { ... }
> struct Type(T)(T x) if (complicated_condition) { ... }
>
> If complicated_condition is not met, the symbol simply 
> disappears and the compiler error message just lists is as a 
> possible, but not viable, candidate.
>
> I think one simple step toward improving things is pushing the 
> condition in a static_assert inside type definitions:
>
> void fun(T)(T x) if (complicated_condition) { ... } // no change
> struct Type(T)(T x)
> {
>   static assert(complicated_condition, "Informative message.");
>   ...
> }
>
> This should improve error messages for types (only). The 
> rationale is that it's okay for types to refuse compilation 
> because types, unlike functions, don't overload. The major 
> reason for template constraints in functions is allowing for 
> good overloading.
>
>
> Andrei

I agree, but also for function should conditions that do not 
effect overloading go into static asserts.

For example if I were to write an algorithm that works with 
forward ranges and can be optimized for random access ranges but 
needs assignable elements in any case:

void foo(R)(R r) if(isForwardRange!R && !isRandomAccessRange!R)
{
     static assert(hasAssignableElements!R, "informative error 
msg");
}

void foo(R)(R r) if(isRandomAccessRange!R)
{
     static assert(hasAssignableElements!R, "informative error 
msg");
}