On Fri, Apr 21, 2023 at 01:35:43AM +0200, Alejandro Colomar wrote:
> On 4/20/23 21:37, наб wrote:
> > diff --git a/man3/regex.3 b/man3/regex.3
> > index 3f1529583..e3dd72a74 100644
> > --- a/man3/regex.3
> > +++ b/man3/regex.3
> > @@ -225,12 +225,10 @@ .SS Error reporting
> > .IR errbuf ;
> > the error string is always null-terminated, and truncated to fit.
> > .SS Freeing
> > -Supplying
> > .BR regfree ()
> > -with a precompiled pattern buffer,
> > -.IR preg ,
> > -will free the memory allocated to the pattern buffer by the compiling
> > -process,
> > +invalidates the pattern buffer at
> While this ("invalidates") is true, it omits the most important information:
> it frees the object.
It doesn't.
> I think it's better to say that it frees (or
> deallocates) the object and any memory allocated within it, since that
> already implies invalidating it (due to
> <https://port70.net/~nsz/c/c11/n1570.html#6.2.4p2> and
> <https://port70.net/~nsz/c/c11/n1570.html#7.22.3p1>),
For the precise reasons listed here:
the regex_t object continues to exist.
regcomp() doesn't allocate *preg, and regfree() doesn't deallocate it.
> and also tells why
> it's necessary to call this function. Otherwise, it's not clear why we
> should call it. Why would I want to invalidate a buffer?
Admittedly, it does also "free any memory allocated by regcomp( )
associated with preg." (Issue 8 Draft 2.1), yeah.
Maybe it's my neurosis that I consider "may no longer be passed to
regexec()" the primary effect here.
Updated to
regfree() invalidates the pattern buffer at *preg, freeing any
associated memory; *preg must have been initialized via regcomp().
Best,
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