function return pointer of int?

[Davy]

Hi all,

I am writing a function, which return the pointer of the int. But it
seems to be wrong. Any suggestion?

int * get_p_t(int t) {
return &t;
}

int main()
{
printf("v1\n");
int t = 5;
int * p_t[2];

p_t[0] = &t; // right
p_t[1] = get_p_t(t); //wrong

return 0;
}

Best regards,
Davy

[James Kanze]

On Nov 6, 3:44*pm, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

James Kanze <james.ka...@gm ail.comwrites:

On Nov 6, 7:39 am, Barry Schwarz <schwa...@dqel. comwrote:

On Wed, 5 Nov 2008 22:30:03 -0800 (PST), baichuan0...@16 3.com wrote:

* * [...]

void *fun(int i)
{
* *void *p = &i;
* *p += (1<<5) + 4;

* *return p;
}

What do you think this accomplishes?

Your first executable statement contains a constraint violation.

I'm not sure what you mean by the "first executable
statement". The definition of p (with its initialization)
definitely generates executable code, and is executed. *And
there's no problem with this statement; it is legal and well
defined, and must work in any implementation.

The cross-post is confusing matters (as always!).

Especially as it concerns an area in which C and C++ try to be
identical---all too often using different words to (hopefully)
say the same thing.

In C there is a clear distinction between declarations and
statements, both in the formal syntax as well as in the less
formal text of the standard, so "the first statement" refers
unambiguously to the increment of p (and the extra
"executable " is redundant). *In C++, as you know, there is a
"declaratio n statement" so the distinction is lost. *Barry is
presumably assuming the code is C.

I'd missed that difference. Historically, of course, it made
sense, since you couldn't mix declarations and (other)
statements.

My main point still holds, of course: no C or C++ compiler will
accept pointer arithmetic on an incomplete type (and void is an
incomplete type). The code simply won't compile.

--
James Kanze (GABI Software) email:ja******* **@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[jameskuyper]

James Kanze wrote:
....

My main point still holds, of course: no C or C++ compiler will
accept pointer arithmetic on an incomplete type (and void is an
incomplete type). The code simply won't compile.

More accurately: no fully conforming compiler for either language will
accept such code without first issuing a diagnostic. A number of
popular compilers will indeed accept pointer arithmetic on void*,
which is performed as if it were char*. As long as they also issue the
mandatory diagnostic message, they can do so while remaining fully
conforming.

[Keith Thompson]

Michael <mi*****@michae ldadmum.no-ip.orgwrites:
[...]

change
int * get_p_t(int t) {
to
int * get_p_t(int&t) {

Not in C (note the cross-post).

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

James Kanze <ja*********@gm ail.comwrites:
[...]

My main point still holds, of course: no C or C++ compiler will
accept pointer arithmetic on an incomplete type (and void is an
incomplete type). The code simply won't compile.

Any conforming C or C++ compiler must issue a diagnostic for any
attempt to peform pointer arithmetic on a pointer to an incomplete
type, since it's a constraint violation. At least in C, it's not
actually required to reject the translation unit; it may legally
compile it successfully, and the resulting program has behavior that
is not defined by the standard (though of course it may be defined by
the implementation) .

The only case where a C translation unit *must* be rejected is when it
contains a #error directive that survives preprocessing. I'm not sure
what the corresponding rules are for C++.

In particular, gcc (specifically the C compiler that's part of the gcc
suite) allows arithmetic on void* by default. This is a permissible
extension as long as it issues a diagnostic. g++ doesn't support this
particular extension by default; I don't know whether it has an option
to enable it.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Michael]

Davy wrote:

Hi all,

I am writing a function, which return the pointer of the int. But it
seems to be wrong. Any suggestion?

int * get_p_t(int t) {
return &t;
}

int main()
{
printf("v1\n");
int t = 5;
int * p_t[2];

p_t[0] = &t; // right
p_t[1] = get_p_t(t); //wrong

return 0;
}

Best regards,
Davy

change
int * get_p_t(int t) {
to
int * get_p_t(int&t) {