Re: do { quit; } else { }

On Thu, 10 Apr 2025 00:20:29 -0000 (UTC)
Kaz Kylheku <643-408-1753@kylheku.com> wrote:

On 2025-04-09, bart <bc@freeuk.com> wrote:

I'm not sure what your gripe is other than maybe I picked up on
something you got wrong. The discussion was about two struct types
like this:
>
    typedef struct tag1 {...} T1;
    typedef struct tag2 {...} T2;
>
and whether T1 and T2 were compatible or not. You said:
>
"and those types are not compatible, because the two struct tags
are different."
>
In this case the tags would be "tag1" and "tag2". I then said:
>
"I get an incompatible error (from the example you snipped) even
when I remove both struct tags." 

 
When you remove the tag from a struct definition, the implementation
behaves as if it were implementing a unique tag which is different
from any other such tag, and any tag that can possibly be written
using textual syntax.
 
How did you implement tagless struct declarations in your compiler?
 

That means removing "tag1" and "tag2" so the example above looks
like this:
>
    typedef struct {...} T1;
    typedef struct {...} T2;
>
Here, you can't say the struct tags are different, as they are not
visible! 

 
So if you close your eyes, two things that were different are now
no longer different, since they are invisible?
 
The tag is a property of the type, not of printed type declaration.
 
A struct type has a tag. If the declaration doesn't show one,
that doesn't mean it doesn't have a tag.
 

In the message above Bart demonstrated that type of struct with
explicit tag depends not just on tag and fields, but also on
lexical scope.
IMHO, instead of going into irrelevant details, it is much simpler to
postulate that all struct declarations within given translation unit
have different types. Upthread, somebody said that already. May be, you.

If a "Simulation" object has a "gravity" member, do you conclude
that a given simulation has no gravity, because the constructor
omitted specifying it?
 
  Simulation s = new Simulation(windSpeed = 35.7)
 

I am not quite understand how C++ is relevant.

As I concluded, your assertion about compatibility being based on
tags being the same or not didn't seem right.) 

 
Or, you know, you could stop caring about what someone wrote in
comp.lang.c, be they right or wrong, and ... look it up?