Is it implementation-defined whether this errors?

[zaimoni]

This is a test case I constructed for a vaporware standalone C99
preprocessor:

#if 1
#elif
#endif

The text did not permit quickly deciding whether the syntax relaxation
for preprocessing directives in blocks that aren't processed (C99 6.10
paragraph 4) dominates knowing what an elif-block actually was (C99
6.10 paragraph 1). If 6.10 paragraph 4 dominates, this should warn
(as the preprocessor got lucky). If C99 6.10 paragraph 1 dominates,
this should error.

The obvious complementary test case

#if 0
#elif
#endif

must error, as the control expression of the #elif is actually
needed. So I tentatively decided to make the first test case behave
like the obvious one: error [why should it only warn just because we
got lucky?].

When I ran my test suite against GCC's cpp, the first test case only
warned. Since this is a legitimate implementation: did I miss
something that guarantees that the syntax relaxation dominates knowing
that the elif-block is well-formed?

[Eric Sosman]

za*****@zaimoni.com wrote:

This is a test case I constructed for a vaporware standalone C99
preprocessor:

#if 1
#elif
#endif

The text did not permit quickly deciding whether the syntax relaxation
for preprocessing directives in blocks that aren't processed (C99 6.10
paragraph 4) dominates knowing what an elif-block actually was (C99
6.10 paragraph 1). If 6.10 paragraph 4 dominates, this should warn
(as the preprocessor got lucky). If C99 6.10 paragraph 1 dominates,
this should error.

The question boils down to whether #elif by itself is a malformed
#elif directive (because there's no expression) or whether it's just
a non-directive (because it doesn't parse as a directive). 6.10p3
settles the issue:

"[...] A non-directive shall not begin with any of the
directive names appearing in the syntax."

That is, no line beginning with #elif can be a non-directive, so it
must be treated as a directive, and since an #elif alone does not
match the syntax of any preprocessor directive it's an error.

The obvious complementary test case

#if 0
#elif
#endif

must error, as the control expression of the #elif is actually
needed.

I see no difference between this example and the first one.

--
Er*********@sun.com

[zaimoni]

On Aug 27, 9:26 am, Eric Sosman <[e-mail]:

[e-mail] wrote:

This is a test case I constructed for a vaporware standalone C99
preprocessor:

#if 1
#elif
#endif

The text did not permit quickly deciding whether the syntax relaxation
for preprocessing directives in blocks that aren't processed (C99 6.10
paragraph 4) dominates knowing what an elif-block actually was (C99
6.10 paragraph 1). If 6.10 paragraph 4 dominates, this should warn
(as the preprocessor got lucky). If C99 6.10 paragraph 1 dominates,
this should error.

The question boils down to whether #elif by itself is a malformed
#elif directive (because there's no expression) or whether it's just
a non-directive (because it doesn't parse as a directive). 6.10p3
settles the issue:

"[...] A non-directive shall not begin with any of the
directive names appearing in the syntax."

That is, no line beginning with #elif can be a non-directive, so it
must be treated as a directive, and since an #elif alone does not
match the syntax of any preprocessor directive it's an error.

Yes -- *if* the #elif directive is being preprocessed.

The obvious complementary test case

#if 0
#elif
#endif

must error, as the control expression of the #elif is actually
needed.

I see no difference between this example and the first one.

Superficially:
Whether the first case's malformed #elif directive is actually being
preprocessed (as a top-level construct) is a "this sentence is false"
paradox. (If I'm going off-course, this is why.) With the second
case, it's a "this sentence is true" non-paradox.

We have to preprocess the #elif directive to determine that there is
an elif-block to not preprocess. (I chose to error at that point,
then discard the malformed elif-block.) But as the starting line of
the not-preprocessed elif-block we're not supposed to preprocess it
after all -- which trips the "any tokens allowed" override for blocks
that aren't preprocessed.

[Eric Sosman]

za*****@zaimoni.com wrote:

On Aug 27, 9:26 am, Eric Sosman <[e-mail]:

>[e-mail] wrote:

>>This is a test case I constructed for a vaporware standalone C99
preprocessor:
#if 1
#elif
#endif
The text did not permit quickly deciding whether the syntax relaxation
for preprocessing directives in blocks that aren't processed (C99 6.10
paragraph 4) dominates knowing what an elif-block actually was (C99
6.10 paragraph 1). If 6.10 paragraph 4 dominates, this should warn
(as the preprocessor got lucky). If C99 6.10 paragraph 1 dominates,
this should error.

The question boils down to whether #elif by itself is a malformed
#elif directive (because there's no expression) or whether it's just
a non-directive (because it doesn't parse as a directive). 6.10p3
settles the issue:

"[...] A non-directive shall not begin with any of the
directive names appearing in the syntax."

That is, no line beginning with #elif can be a non-directive, so it
must be treated as a directive, and since an #elif alone does not
match the syntax of any preprocessor directive it's an error.

Yes -- *if* the #elif directive is being preprocessed.

It is a directive, and it is not inside a skipped group, so
it is processed. The grammar of 6.10p1 makes it clear that the
#elif is not part of the if-group that is included or omitted by
the #if, but marks the start of the elif-group that follows the
if-group in the same if-section. Thus, 6.10.1p5 does not apply.

>>The obvious complementary test case
#if 0
#elif
#endif
must error, as the control expression of the #elif is actually
needed.

I see no difference between this example and the first one.

Superficially:
Whether the first case's malformed #elif directive is actually being
preprocessed (as a top-level construct) is a "this sentence is false"
paradox. (If I'm going off-course, this is why.) With the second
case, it's a "this sentence is true" non-paradox.

I'm also susceptible to going off-course, but you've not yet
convinced me that the course and I have become non-proximal ...
As I understand the matter, the #elif directive *is* processed, no
matter what the value of the controlling expression in the preceding
#if or #elif, because the #elif is not part of the group that the
preceding directive includes or suppresses.

6.10.1p5 applies to situations like this, which I think is
valid C despite its unusual appearance:

#if 0
#if *-*-*-*-*-*-*-*
#elif |-|-|-|-|-|-|-|
#elif <<<<<-=-=->>>>>
#endif you must this old grey head
#elif 0
#else
#endif

IMHO, the directives in the inner if-section -- fully contained within
the if-group of the outer if-section -- are processed only as far as
their names, as 6.10.1p5 says. The other four directives are, I think,
fully processed regardless of the control expressions' values.

--
Er*********@sun.com

[zaimoni]

On Aug 27, 12:32 pm, Eric Sosman <[e-mail]:

[e-mail] wrote:

On Aug 27, 9:26 am, Eric Sosman <[e-mail]:

[e-mail] wrote:
This is a test case I constructed for a vaporware standalone C99
preprocessor:
#if 1
#elif
#endif

>....

That is, no line beginning with #elif can be a non-directive, so it
must be treated as a directive, and since an #elif alone does not
match the syntax of any preprocessor directive it's an error.

Yes -- *if* the #elif directive is being preprocessed.

It is a directive, and it is not inside a skipped group, so
it is processed. The grammar of 6.10p1 makes it clear that the
#elif is not part of the if-group that is included or omitted by
the #if, but marks the start of the elif-group that follows the
if-group in the same if-section.

Agreed.

Thus, 6.10.1p5 does not apply.

6.10.1p5 entails the tie-breaker I needed, however.

>The obvious complementary test case
#if 0
#elif
#endif
must error, as the control expression of the #elif is actually
needed.
I see no difference between this example and the first one.

Superficially:
Whether the first case's malformed #elif directive is actually being
preprocessed (as a top-level construct) is a "this sentence is false"
paradox. (If I'm going off-course, this is why.) With the second
case, it's a "this sentence is true" non-paradox.

What I had not noticed: 6.10.1p5 entails that an #elif directive does
not suppress itself. This dispels the claimed paradox. (I plausibly
overlooked a more direct statement to that effect.)

....

6.10.1p5 applies to situations like this, which I think is
valid C despite its unusual appearance:

I'm certain your following example is valid C.

#if 0
#if *-*-*-*-*-*-*-*
#elif |-|-|-|-|-|-|-|
#elif <<<<<-=-=->>>>>
#endif you must this old grey head
#elif 0
#else
#endif

IMHO, the directives in the inner if-section -- fully contained within
the if-group of the outer if-section -- are processed only as far as
their names, as 6.10.1p5 says.

Yes, this is unequivocal.