const char problem

[sugaray]

I wrote a short code as shown below for experiment purpose,
It is successfully compilable, yet I found the piece I wrote
even confused myself with the consts and the asterisks, and
which brought up three questions to me:

1) will the casting in line 1 affect the rest of the
declarations ?
2) what's the difference between line 2 and line 3
3) are there any practical purposes of the usage of any
of the three ?

#include <stdio.h>

int main(int argc,char **argv)
{

const char **pa=(const char **)argv; // line 1
const char *const *p=argv; // line 2
const char *const *const pp=argv; // line 3

printf("%s %c\n",*++p,**p) ;
printf("%s %c\n",*pp,**pp) ;

return 0;
}

[pete]

sugaray wrote:

I wrote a short code as shown below for experiment purpose,
It is successfully compilable, yet I found the piece I wrote
even confused myself with the consts and the asterisks, and
which brought up three questions to me:

1) will the casting in line 1 affect the rest of the
declarations ?
No, not even if the cast had any effect at all, which it doesn't.
2) what's the difference between line 2 and line 3
pp has a const qulaifier that p doesn't have.
3) are there any practical purposes of the usage of any
of the three ?

#include <stdio.h>

int main(int argc,char **argv)
{

const char **pa=(const char **)argv; // line 1
const char *const *p=argv; // line 2
const char *const *const pp=argv; // line 3

printf("%s %c\n",*++p,**p) ;
The above line is unspecified behavior
depending on which argument is evaluated first.
printf("%s %c\n",*pp,**pp) ;

return 0;
}

--
pete

[Dan Pop]

In <40***********@ mindspring.com> pete <pf*****@mindsp ring.com> writes:

printf("%s %c\n",*++p,**p) ;

The above line is unspecified behavior
depending on which argument is evaluated first.

Nope, it's undefined behaviour, because the old value of p is used for
other purposes than computing its new value, with no intervening
sequence point.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Richard Bos]

pete <pf*****@mindsp ring.com> wrote:

sugaray wrote:

printf("%s %c\n",*++p,**p) ;

The above line is unspecified behavior
depending on which argument is evaluated first.

Undefined. p gets read and modified without a sequence point in between.

Richard

[Dan Pop]

In <40************ ***@news.indivi dual.net> rl*@hoekstra-uitgeverij.nl (Richard Bos) writes:

pete <pf*****@mindsp ring.com> wrote:

sugaray wrote:

> printf("%s %c\n",*++p,**p) ;

The above line is unspecified behavior
depending on which argument is evaluated first.

Undefined. p gets read and modified without a sequence point in between.

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^
If that were enough for invoking undefined behaviour, i = i + 1 would
invoke undefined behaviour. The actual rule is slightly more complex.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[pete]

Dan Pop wrote:

In <40***********@ mindspring.com> pete <pf*****@mindsp ring.com> writes:

printf("%s %c\n",*++p,**p) ;

The above line is unspecified behavior
depending on which argument is evaluated first.

Nope, it's undefined behaviour, because the old value of p is used for
other purposes than computing its new value, with no intervening
sequence point.

Thank you.

--
pete