why is ifThrown un at safe?

[Bastiaan Veelo]

On Friday, 15 March 2019 at 18:46:25 UTC, bauss wrote:
> On Friday, 15 March 2019 at 18:04:05 UTC, Bastiaan Veelo wrote:
>> In the code below (https://run.dlang.io/is/d0oTNi), ifThrown 
>> is inferred as un at safe. If instead I write the implementation 
>> of ifThrown out (after res2) then it is @safe. As far as I can 
>> see, there is no real difference. So why doesn't ifThrown work 
>> in this case, and can it be made to work?
>>
>> Thanks!
>>
>> void main() @safe
>> {
>>     import std.process;
>>     import std.exception;
>>
>>      const res1 = execute(["clang", "-v", "-xc++", 
>> "/dev/null", "-fsyntax-only"], ["LANG": "C"])
>>         .ifThrown((e) @safe {
>>             import std.typecons : Tuple;
>>             return Tuple!(int, "status", string, "output")(-1, 
>> e.msg);
>>         }); // Fails
>>
>>     const res2 = () {
>>         try
>>         {
>>             return execute(["clang", "-v", "-xc++", 
>> "/dev/null", "-fsyntax-only"], ["LANG": "C"]);
>>         }
>>         catch (Exception e)
>>         {
>>             import std.typecons : Tuple;
>>             return Tuple!(int, "status", string, "output")(-1, 
>> e.msg);
>>         }
>>     }();
>> }
>
> Because the handlers may be unsafe.

But this handler is explicitly marked @safe, and ifThrown is a 
template, which should infer its attributes, right?

> There is no safe overload of ifThrown.

Could it be added?

> However you can work around this using @trusted.

I know, but since everything really is safe, I'd rather not imply 
that it might not be. Besides, I failed to use @trusted without 
introducing a new scope, so I'd have to mark the whole block 
where `res1` is used as @trusted. Maybe I did that wrong though.