pointer concepts

[Newbie]

Hi,

I'm a newbie. I need to clear my pointer concepts.

the code below gives segmentation fault when i use char* str;
int main()
{
char *str;
printf("Enter a string to reverse\n");
scanf("%s",&str );
strlen(str);
}

if I use char str[20] it runs fine.
int main()
{
char str[20];
printf("Enter a string to reverse\n");
scanf("%s",&str );
strlen(str);
}

Can you please explain why?

Thanks

[Nelu]

Newbie wrote:

Hi,

I'm a newbie. I need to clear my pointer concepts.

the code below gives segmentation fault when i use char* str;
int main() use: int main(void) {
char *str;
printf("Enter a string to reverse\n");
scanf("%s",&str ); Here's the problem. str points to somewhere in memory
but you have no idea where and whether or not that area
belongs to you. You need to allocate memory (dynamically).
You can run into the problem again if you read more
characters than you allocated. strlen(str); return something. }

if I use char str[20] it runs fine. Well, yes, because you str[20] is automatically
allocated and you tell it exactly how much memory
you want. To dynamically allocate memory you need
to use a pointer and allocate memory for it (malloc & co). int main() int main(void) {
char str[20];
printf("Enter a string to reverse\n");
scanf("%s",&str );
strlen(str); return something }

Can you please explain why?

That's it.
--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Vladimir S. Oka]

Nelu wrote:

Newbie wrote:

int main() use: int main(void)

{
char *str;
printf("Enter a string to reverse\n");
scanf("%s",&str );

Here's the problem. str points to somewhere in memory
but you have no idea where and whether or not that area
belongs to you. You need to allocate memory (dynamically).
You can run into the problem again if you read more
characters than you allocated.

Another problem, as big as not allocating any space for the string, is
that scanf("%s",...) requires `char*`, and you give it `char**`. You
(OP) used `&str` which yields the address of `str` variable, /not/ the
string it points to (even if one exists).

You don't seem to have high enough compiler warning setting, or you
ignored the one you got. My gcc complained...

The badness can be masked also if you just quickly test with the string
that will fit into the sizeof(char*) on your system (it would break
later, when you try to use now broken pointer).

strlen(str);

return something.

}

if I use char str[20] it runs fine.

Well, yes, because you str[20] is automatically
allocated and you tell it exactly how much memory
you want. To dynamically allocate memory you need
to use a pointer and allocate memory for it (malloc & co).

int main()

int main(void)

{
char str[20];
printf("Enter a string to reverse\n");
scanf("%s",&str );
strlen(str);

return something

}

Can you please explain why?

Here, you have exactly the same problem as above. Use `str` and not
`&str` for scanf(). Don't get confused by other types. Revisit your
textbook sections on pointers and C strings.
That's it.

Indeed.

--
BR, Vladimir

"OK, now let's look at four dimensions on the blackboard."
-- Dr. Joy

[geshule]

elu wrote:

You can run into the problem again if you read more
characters than you allocated.

I agree with you on that.
But the codes below works even if you input more than 10 characters?
Why?

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int main()
{
char *str;
str = new char[10];
printf("Enter a string to reverse\n");
scanf("%s",str) ;
cout << strlen(str)<< endl;
system("pause") ;
return 0;
}

[Vladimir S. Oka]

ge*****@hotmail .com wrote:

elu wrote:

You can run into the problem again if you read more
characters than you allocated.

I agree with you on that.
But the codes below works even if you input more than 10 characters?
Why?

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int main()
{
char *str;
str = new char[10];
printf("Enter a string to reverse\n");
scanf("%s",str) ;
cout << strlen(str)<< endl;
system("pause") ;
return 0;
}

I don't know, it's not C!
Please post C++ questions to comp.lang.c++

<OT>
I don't see why would it work for more characters than you allocate.
</OT>

--
BR, Vladimir

Goto, n.:
A programming tool that exists to allow structured programmers
to complain about unstructured programmers.
-- Ray Simard

[Nelu]

Vladimir S. Oka wrote:

Nelu wrote:

Newbie wrote:

int main()

use: int main(void)

{
char *str;
printf("Enter a string to reverse\n");
scanf("%s",&str );

Here's the problem. str points to somewhere in memory
but you have no idea where and whether or not that area
belongs to you. You need to allocate memory (dynamically).
You can run into the problem again if you read more
characters than you allocated.

Another problem, as big as not allocating any space for the string, is
that scanf("%s",...) requires `char*`, and you give it `char**`. You
(OP) used `&str` which yields the address of `str` variable, /not/ the
string it points to (even if one exists).

You don't seem to have high enough compiler warning setting, or you
ignored the one you got. My gcc complained...

The badness can be masked also if you just quickly test with the string
that will fit into the sizeof(char*) on your system (it would break
later, when you try to use now broken pointer).

Right, I forgot to tell him that, and that if he uses an array
& will have the same address and wouldn't crash the program.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Nelu]

ge*****@hotmail .com wrote:

elu wrote:

You can run into the problem again if you read more
characters than you allocated.

I agree with you on that.
But the codes below works even if you input more than 10 characters?
Why?

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int main()
{
char *str;
str = new char[10];
printf("Enter a string to reverse\n");
scanf("%s",str) ;
cout << strlen(str)<< endl;
system("pause") ;
return 0;
}

This is not C. It is C++.
Also, some OSes (or kernel patches) try to avoid buffer overflows
and won't crash the program. However, don't trust them. It's wrong.
I had the same problem with someone who wrote a shell on RedHat a few
years ago and it was working just fine on his machine but crashed
instantly on my slackware. When I checked the code I was stunned.
He was using pointers without allocating any memory for them. His
kernel was patched to deal with buffer overflows so, even if his
program was way wrong it was not crashing. That's not normal behavior.
--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Vladimir S. Oka]

Nelu wrote:

Also, some OSes (or kernel patches) try to avoid buffer overflows
and won't crash the program. However, don't trust them. It's wrong.

This sounds absolutely /horrible/!

--
BR, Vladimir

Song Title of the Week:
"They're putting dimes in the hole in my head to see the change
in me."

[Richard Heathfield]

Nelu said:

Vladimir S. Oka wrote:

Another problem, as big as not allocating any space for the string, is
that scanf("%s",...) requires `char*`, and you give it `char**`. You
(OP) used `&str` which yields the address of `str` variable, /not/ the
string it points to (even if one exists).

You don't seem to have high enough compiler warning setting, or you
ignored the one you got. My gcc complained...

The badness can be masked also if you just quickly test with the string
that will fit into the sizeof(char*) on your system (it would break
later, when you try to use now broken pointer).

Right, I forgot to tell him that, and that if he uses an array
& will have the same address and wouldn't crash the program.

Quibble: given char arr[20], arr (in a value context) has type char *,
whereas &arr has type char (*)[20] - these are different and incompatible
types.

More importantly: what nobody seems to have mentioned is that scanf("%s" is
a remarkably dangerous construct, in that it offers no buffer protection
whatsoever.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

[Nelu]

Richard Heathfield wrote:

Quibble: given char arr[20], arr (in a value context) has type char *,
whereas &arr has type char (*)[20] - these are different and incompatible
types. Yes. This was discussed on the list recently, I think.
More importantly: what nobody seems to have mentioned is that scanf("%s" is
a remarkably dangerous construct, in that it offers no buffer protection
whatsoever.

I think I told the OP about the problem if the buffer overflows.
--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)