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]

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) ?

 

[kongweize]

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]

(long *) instead of (* long) ?

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

 

[vippstar]

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]

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]

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

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.

 

[Peter Nilsson]

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 *).

 

[Ian Collins]

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.