I think this is a problem in gcc.

[Aereshaa]

For some reason, when I compile this code:

int main(){
char* a = malloc(5);
long* l = (*long) a;
} //I shortened it to isolate the problem.

I get this error:

error.c:3: error: expected expression before ‘long’

But there IS an expression before 'long'. Why is this?

[Chris McDonald]

Aereshaa <Ae******@gmail.comwrites:

>For some reason, when I compile this code:

>int main(){
char* a =3D malloc(5);
long* l =3D (*long) a;
} //I shortened it to isolate the problem.

>I get this error:

>error.c:3: error: expected expression before =91long=92

>But there IS an expression before 'long'. Why is this?

(long *) instead of (* long) ?

--
Chris.

[kongweize]

On Jun 2, 11:38*am, Aereshaa <[e-mail]:

For some reason, when I compile this code:

int main(){
*char* a = malloc(5);
*long* l = (*long) a;

} //I shortened it to isolate the problem.

I get this error:

error.c:3: error: expected expression before ‘long’

But there IS an expression before 'long'. Why is this?

I think you made mistake in
long* l = (*long) a;

the * in (*long) is recognized as a multiplication sign

[Aereshaa]

On Jun 1, 11:42 pm, Chris McDonald <[e-mail]:

Aereshaa <[e-mail]:

For some reason, when I compile this code:
int main(){
char* a =3D malloc(5);
long* l =3D (*long) a;
} //I shortened it to isolate the problem.
I get this error:
error.c:3: error: expected expression before =91long=92
But there IS an expression before 'long'. Why is this?

(long *) instead of (* long) ?

--
Chris.

Oh, #it! I was under the impression that it referred to the first
'long' in the line.
Thanks.

[vippstar]

On Jun 2, 6:48 am, Aereshaa <[e-mail]:

On Jun 1, 11:42 pm, Chris McDonald <[e-mail]:

Aereshaa <[e-mail]:

>For some reason, when I compile this code:
>int main(){
char* a =3D malloc(5);
long* l =3D (*long) a;
>} //I shortened it to isolate the problem.
>I get this error:
>error.c:3: error: expected expression before =91long=92
>But there IS an expression before 'long'. Why is this?

(long *) instead of (* long) ?

--
Chris.

Oh, #it! I was under the impression that it referred to the first
'long' in the line.
Thanks.

What you probably want:
l = *(long *)a
However sizeof (long) is not guaranteed to be less or equal to 5, and
the contents of a are not initialized.
Try this instead:
char *a = malloc(sizeof (long)):
long l;
if(a) {
memset(a, 0, sizeof(long));
l = *(long*)a;
}

[Barry Schwarz]

On Sun, 1 Jun 2008 20:48:08 -0700 (PDT), Aereshaa <Ae******@gmail.com>
wrote:

>On Jun 1, 11:42 pm, Chris McDonald <[e-mail]:

>Aereshaa <[e-mail]:

>For some reason, when I compile this code:
int main(){
char* a =3D malloc(5);
long* l =3D (*long) a;
} //I shortened it to isolate the problem.
I get this error:
error.c:3: error: expected expression before =91long=92
But there IS an expression before 'long'. Why is this?

(long *) instead of (* long) ?

--
Chris.

Oh, #it! I was under the impression that it referred to the first
'long' in the line.
Thanks.

After fixing the cast as suggested, be aware that if sizeof(long)
exceeds 5 than any attempt to dereference l yields undefined behavor.
The same is true for a if sizeof(int) exceeds 5 but that is less
likely.
Remove del for email

[Keith Thompson]

vi******@gmail.com writes:

On Jun 2, 6:48 am, Aereshaa <[e-mail]:

>On Jun 1, 11:42 pm, Chris McDonald <[e-mail]:

Aereshaa <[e-mail]:
For some reason, when I compile this code:
int main(){
char* a = malloc(5);
long* l = (*long) a;
} //I shortened it to isolate the problem.
I get this error:
error.c:3: error: expected expression before 'long'
But there IS an expression before 'long'. Why is this?

(I fixed some quoted-printable noise in the above.)

(long *) instead of (* long) ?

Oh, #it! I was under the impression that it referred to the first
'long' in the line.
Thanks.

What you probably want:
l = *(long *)a

I doubt it, unless he's going to change the code substantially.
``l'' is of type long*, not long.

However sizeof (long) is not guaranteed to be less or equal to 5, and
the contents of a are not initialized.
Try this instead:
char *a = malloc(sizeof (long)):
long l;
if(a) {
memset(a, 0, sizeof(long));
l = *(long*)a;
}

That's a rather roundabout way of writing ``long l = 0;'', which
doesn't particularly resemble what the OP was trying to do.

But the OP did say that the posted code was shortened to isolate the
problem. It's not surprising that a piece of code intended only to
exhibit a syntax error wouldn't make much sense logically.

If the OP really is trying to assign the result of malloc(5) to a
long*, I hope he(?) will post again and learn why it's a bad idea.

--
Keith Thompson (The_Other_Keith) 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"

[Peter Nilsson]

"[e-mail]" <[e-mail]:

Aereshaa wrote:

error.c:3: error: expected expression before ‘long’

I think you made mistake in
long* l = (*long) a;

the * in (*long) is recognized as a multiplication sign

Not that it matters, but I think it's much more likely
it's recognised as the indirection operator (unary *).

--
Peter

[Ian Collins]

Aereshaa wrote:

For some reason, when I compile this code:

int main(){
char* a = malloc(5);
long* l = (*long) a;
} //I shortened it to isolate the problem.

I get this error:

error.c:3: error: expected expression before ‘long’

But there IS an expression before 'long'. Why is this?

Ignoring the cast problem, this would be more than a little dangerous on
any IPL64 system.

--
Ian Collins.