Re: int a = a (Was: Bart's Language)

On 2025-03-19, David Brown <david.brown@hesbynett.no> wrote:

On 18/03/2025 19:36, Janis Papanagnou wrote:

On 18.03.2025 19:04, Kenny McCormack wrote:

In article <vrc75b$2r4lt$1@dont-email.me>,
David Brown  <david.brown@hesbynett.no> wrote:
...

gcc won't warn until you say '-Wextra', and then only for:
>
    int a = a + 1;

>
People would not normally write "int a = a;".  It is used as a common
idiom meaning "I know it is not clear to the compiler that the variable
is always initialised before use, but /I/ know it is - so disable the
use-without-initialisation warnings for this variable".

 
Wow! - It would never have occurred to me that "int a = a;" being
considered an idiom, let alone a "common idiom".
 

>
It is certainly an idiom, and certainly viewed by gcc as a way to avoid
an "uninitialized" warning (unless "-Winit-self" is also enabled), and
it is an idiom I have seen documented in at least one other compiler
(though I can't remember which - I've read many compiler manuals over
the decades).

I think that this "idiom" is a complete fluke, except in that compiler
where it was documented.

It's easy to see how it can arise "naturally" in a compiler which has
warnings about both uses of uninitialized variables, and about
variables being left uninitialized.

In the "int a = a;" declaration, the initializing expression "a"
is in the scope of a.

Suppose the compiler, when processing the declaration, begins
by extending the compile-time scope object with the name "a",
and seeing that an initializer is present in the syntax, it
flags "a" as being initialized. That suppresses any diagnostics
about a not being initialized.

Then it subsequently processes the initializing expression "a".
At that point "a" is flagged initialized, and so no diagnostic
occurs for the use of uninitialized "a".

That's a bug in the compiler, not an idiom. "a" is positively *not*
initialized until the initializing expression is evaluated.
The compiler must *not* set this "initialized" flag in the
scope information until it has compiled the initializing expression
and moved on to the next declarator, declaration or statement.
That way when it is code-walking the initializing expression,
it will encounter "a" and correctly report that it's being
accessed without having been initialized.

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