why debug step by step, it s ok

[baumann]

Hi all,
I have one simple program:
#define __EXTENSIONS__
#include <stdio.h>
#include <string.h>
int main() {
char *buf="5/90/45";
char *token;
char *lasts;
printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) {
printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

the program has abused the strok, i know it,

******but why when i `gdb prog_name` and execute step by step , it 's
ok?
thanks

baumann@pan

[Mark F. Haigh]

baumann@pan wrote:

Hi all,

<code snipped>
the program has abused the strok, i know it,

******but why when i `gdb prog_name` and execute step by step , it 's
ok?

Undefined behavior is exactly that: undefined. The code might work
fine, crash, generate nasal demons, or cause mudslides in California.

The point is, you don't know.
Mark F. Haigh
mf*****@sbcglob al.net

[Michael Knaup]

baumann@pan wrote:

Hi all,
I have one simple program:
#define __EXTENSIONS__
#include <stdio.h>
#include <string.h>
int main() {
char *buf="5/90/45"; buf is a pointer to a string constant
char buf[] = "5/90/45"
is correct.
char *token;
char *lasts;
printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) { strtok changes the content of buf, so the content must be changeable.
printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

the program has abused the strok, i know it,

******but why when i `gdb prog_name` and execute step by step , it 's
ok?
thanks

baumann@pan

--
Michael Knaup

[baumann]

you donn't answer my question,

i have answered the question in another post thread.

and in practice, i found when in debug and exec step by step, the prog
runs well.
Michael Knaup wrote:

baumann@pan wrote:

Hi all,
I have one simple program:
#define __EXTENSIONS__
#include <stdio.h>
#include <string.h>
int main() {
char *buf="5/90/45";

buf is a pointer to a string constant
char buf[] = "5/90/45"
is correct.

char *token;
char *lasts;
printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) {

strtok changes the content of buf, so the content must be changeable.

printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

the program has abused the strok, i know it,

******but why when i `gdb prog_name` and execute step by step , it 's
ok?
thanks

baumann@pan

--
Michael Knaup

[baumann]

you can read the post
http://groups-beta.google.com/group/...000d5e177771c8

the 2nd post i pointed out the problem, the 3rd post, i bring forward
the problem asked here.

Michael Knaup wrote:

baumann@pan wrote:

Hi all,
I have one simple program:
#define __EXTENSIONS__
#include <stdio.h>
#include <string.h>
int main() {
char *buf="5/90/45";

buf is a pointer to a string constant
char buf[] = "5/90/45"
is correct.

char *token;
char *lasts;
printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) {

strtok changes the content of buf, so the content must be changeable.

printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

the program has abused the strok, i know it,

******but why when i `gdb prog_name` and execute step by step , it 's
ok?
thanks

baumann@pan

--
Michael Knaup

[Michael Knaup]

baumann@pan wrote:

you can read the post
http://groups-beta.google.com/group/...000d5e177771c8
the 2nd post i pointed out the problem, the 3rd post, i bring forward
the problem asked here.

Michael Knaup wrote:

baumann@pan wrote:

> Hi all,
>
>
> I have one simple program:
> #define __EXTENSIONS__
> #include <stdio.h>
> #include <string.h>
>
>
> int main() {
> char *buf="5/90/45";

buf is a pointer to a string constant
char buf[] = "5/90/45"
is correct.

> char *token;
> char *lasts;
>
>
> printf("tokeniz ing \"%s\" with strtok():\n", buf);
> if ((token = strtok(buf, "/")) != NULL) {

strtok changes the content of buf, so the content must be changeable.

> printf("token = \"%s\"\n", token);
> while ((token = strtok(NULL, "/")) != NULL) {
> printf("token = \"%s\"\n", token);
> }
> }
> }
>
> the program has abused the strok, i know it,
>
> ******but why when i `gdb prog_name` and execute step by step , it 's
> ok?
>
>
> thanks
>
> baumann@pan

--
Michael Knaup

I showed the solution for the problem and M. F. Haigh wrote the
answer to your question.

And here, espacially for you, again:

Modifiying a string literal leads to undefined behavior.
This means it might work in one execution environment (gdb)
or crash, or format your disk, or ... in another.

--
Michael Knaup

[Michael Knaup]

Michael Knaup wrote:

so where is the difference between

(1) char *buffer = "Constant Literal";

and
(2) char buffer[17] = "Constant Literal";

In (1) you define a pointer to char in (2) you define an
array of char's.

In (1) the pointer points to the constant string literal,
in (2) the array is initialised with the char's of the
constant literal. This means (2) is equivalent to
(2) char buffer[17] = { 'C', 'o', ..., 'l', '\000' };

And cause the compiler knows the length of the initialising
string literal in (2) you are not required to give the array
size explicitly so you can just write
char buffer[] = "Con...";
--
Michael Knaup

[Emmanuel Delahaye]

baumann@pan wrote on 03/06/05 :

#include <stdio.h>
#include <string.h>

int main() {
char *buf="5/90/45";
char buf[]="5/90/45";
char *token;
char *lasts;

printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) {
printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

strings are not mutable...

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"Mal nommer les choses c'est ajouter du malheur au
monde." -- Albert Camus.

[Emmanuel Delahaye]

baumann@pan wrote on 03/06/05 :

you donn't answer my question,

i have answered the question in another post thread.

and in practice, i found when in debug and exec step by step, the prog
runs well.

An Undefined Behaviour has an undefined behavour.

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"Mal nommer les choses c'est ajouter du malheur au
monde." -- Albert Camus.

[Robert Gamble]

On Fri, 03 Jun 2005 17:12:39 +0200, Emmanuel Delahaye wrote:

baumann@pan wrote on 03/06/05 :

#include <stdio.h>
#include <string.h>

int main() {
char *buf="5/90/45";

char buf[]="5/90/45";

char *token;
char *lasts;

printf("tokeniz ing \"%s\" with strtok():\n", buf);
if ((token = strtok(buf, "/")) != NULL) {
printf("token = \"%s\"\n", token);
while ((token = strtok(NULL, "/")) != NULL) {
printf("token = \"%s\"\n", token);
}
}
}

strings are not mutable...

Psst: string _literals_ are not mutable.

Robert Gamble