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Answer 1

+10 A:

The end of an initializer is a sequence point so the example in the title is legal.

The comma operator is also a sequence point so that your "normal statement" is also legal and well-defined.

The wikipedia article has a list of the C and C++ sequence points.

To follow up on a comment below, here's a demonstration of the fearsome power of the comma operator, as preserved in FreeBSD's stdio.h (under ifndef __GNUC__):

/*
 * This has been tuned to generate reasonable code on the vax using pcc.
 */
#define __sputc(c, p) \
        (--(p)->_w < 0 ? \
                (p)->_w >= (p)->_lbfsize ? \
                        (*(p)->_p = (c)), *(p)->_p != '\n' ? \
                                (int)*(p)->_p++ : \
                                __swbuf('\n', p) : \
                        __swbuf((int)(c), p) : \
                (*(p)->_p = (c), (int)*(p)->_p++))
#endif

Ben Jackson 2010-10-31 02:50:24

I read the wiki article, but not closely enough to register the comma operator bit. (Cursed inconsistencies.) Anyway, GCC never issued any warnings to me and it always worked, so I kind of assumed it did.

Chris Lutz 2010-10-31 02:57:40

Prior to GCC's statement-expression extension (`({ statements...; value })` the comma operator was the main way to make really complex macros (eg stdio's putc).

Ben Jackson 2010-10-31 03:02:04

@Ben - I would hate to see `putc()` defined that way. That path can only lead to madness.

Chris Lutz 2010-10-31 03:05:22

Not to mention, the idea that inlining that much code for every `putc` is better than making a simple function call is outdated by a good 10-20 years..

R.. 2010-10-31 05:06:24

Answer 2

A:

Thats a tricky question. If you had written:

int i = 0, j = 0, x = 10;
i = x++, j = x++;

I had said: it is undefined. (Various C and C++ standards keep that undfined)

But in an initializer I'm not sure ;D ofc I had concluded it is there undefined as well, but who knows ...

Angelo

Angel O'Sphere 2010-10-31 02:51:29

I was incorrect: the comma operator is a sequence point. (What would we use it for if `f(), g()` where `f()` and `g()` are both `puts(__FUNC__)` had undefined behavior?) My question still stands, but I'm led to believe you're incorrect.

Chris Lutz 2010-10-31 03:04:16

-1 for incorrect answer. Comma operator is a sequence point.

R.. 2010-10-31 05:07:04

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