simple C code

[Saifi Khan-3]

Hi all:

Here is a simple C code.

#include <stdio.h>

int main(int argc, char *argv[])
{
    char *a;

    int **b;

    b = &(int *)a;

    return 0;
}


--
any issues with this piece of code ?


thanks
Saifi.

[uma-8]

Hi Saifi,

The following line should give compiler error
   b = &(int *)a;

'&' operator expects operand immediately, but not typecast operator or any
other expression.

It could be corrected as:

#include <stdio.h>

int main(int argc, char *argv[]) {
char *a;
int **b;
int *c;

c = (int *)a;
b = &c;

return 0;
}

But the operation is not portable.

thanks,
uma..

On Sat, Jul 18, 2009 at 10:06 AM, Saifi Khan <saifi.khan@...>wrote:

> Hi all:
>
> Here is a simple C code.
>
> #include <stdio.h>
>
> int main(int argc, char *argv[])
> {
>    char *a;
>
>    int **b;
>
>    b = &(int *)a;
>
>    return 0;
> }
>
>
> --
> any issues with this piece of code ?
>
>
> thanks
> Saifi.
>

[Non-text portions of this message have been removed]

[Mukund]

Hi uma,

#include <stdio.h>

int main(int argc, char *argv[]) {
int *a;
int **b;
int *c;

c = (int *)a;
b = &c;

return 0;
}

This seems to work why?

thanks,

SMS

On Sat, Jul 18, 2009 at 2:01 PM, uma <uma@...> wrote:

>
>
> Hi Saifi,
>
> The following line should give compiler error
> b = &(int *)a;
>
> '&' operator expects operand immediately, but not typecast operator or any
> other expression.
>
> It could be corrected as:
>
>
> #include <stdio.h>
>
> int main(int argc, char *argv[]) {
> char *a;
> int **b;
> int *c;
>
> c = (int *)a;
> b = &c;
>
> return 0;
> }
>
> But the operation is not portable.
>
> thanks,
> uma..
>
> On Sat, Jul 18, 2009 at 10:06 AM, Saifi Khan <saifi.khan@...<saifi.khan%40twincling.org>
> >wrote:
>
> > Hi all:
> >
> > Here is a simple C code.
> >
> > #include <stdio.h>
> >
> > int main(int argc, char *argv[])
> > {
> > char *a;
> >
> > int **b;
> >
> > b = &(int *)a;
> >
> > return 0;
> > }
> >
> >
> > --
> > any issues with this piece of code ?
> >
> >
> > thanks
> > Saifi.
> >
>
> [Non-text portions of this message have been removed]
>
>  
>

[Non-text portions of this message have been removed]

[Saifi Khan-3]

On Sat, 18 Jul 2009, Mukund Deshpande wrote:

> Hi uma,
>
> #include <stdio.h>
>
> int main(int argc, char *argv[]) {
> int *a;
> int **b;
> int *c;
>
> c = (int *)a;
> b = &c;
>
> return 0;
> }
>
> This seems to work why?
>
> thanks,
>
> SMS
>
> On Sat, Jul 18, 2009 at 2:01 PM, uma <uma@...> wrote:
>
> >
> >
> > Hi Saifi,
> >
> > The following line should give compiler error
> > b = &(int *)a;
> >
> > '&' operator expects operand immediately, but not typecast operator or any
> > other expression.
> >
> > It could be corrected as:
> >
> >
> > #include <stdio.h>
> >
> > int main(int argc, char *argv[]) {
> > char *a;
> > int **b;
> > int *c;
> >
> > c = (int *)a;
> > b = &c;
> >
> > return 0;
> > }
> >
> > But the operation is not portable.
> >
> > thanks,
> > uma..
> >
> > On Sat, Jul 18, 2009 at 10:06 AM, Saifi Khan <saifi.khan@...<saifi.khan%40twincling.org>
> > >wrote:
> >
> > > Hi all:
> > >
> > > Here is a simple C code.
> > >
> > > #include <stdio.h>
> > >
> > > int main(int argc, char *argv[])
> > > {
> > > char *a;
> > >
> > > int **b;
> > >
> > > b = &(int *)a;
> > >
> > > return 0;
> > > }
> > >
> > >
> > > --
> > > any issues with this piece of code ?
> > >
> > >
> > > thanks
> > > Saifi.
> > >
> >

I. compiling with gcc

lval.c: In function 'main':
lval.c:9: error: lvalue required as unary '&' operand


II. compiling with clang/llvm

lval.c:9:9: error: address expression must be an lvalue or a function designator
    b = &(int *)a;
        ^~~~~~~~~
1 diagnostic generated.


thanks
Saifi.