simple problem

[J Wang]

the c code as follows:

#include <stdio.h>
#include <stdlib.h>

int main()
{
printf("%s\n", getenv("NAME"));
return 0;
}

Segmentation Fault

why?

cheers,

[Leor Zolman]

On Fri, 9 Apr 2004 23:51:26 GMT, J Wang <cs****@bath.ac.uk> wrote:

the c code as follows:

#include <stdio.h>
#include <stdlib.h>

int main()
{
printf("%s\n", getenv("NAME"));
return 0;
}

Segmentation Fault

why?
NULL.
-leor

cheers,

--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: Download BD Software's free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html

[Mike Wahler]

"J Wang" <cs****@bath.ac.uk> wrote in message
news:Pi**************************************@amos .bath.ac.uk...

the c code as follows:

#include <stdio.h>
#include <stdlib.h>

int main()
{
char *name = getenv("NAME");
printf("%s\n", getenv("NAME"));
if(name)
printf("%s\n", name);
else
printf("%s\n", "getenv() returned NULL");
return 0;
}

Segmentation Fault

why?

Most likely you tried to dereference a NULL pointer.

*Always* check the return value of a function
which can fail.

-Mike

[Martin Ambuhl]

J Wang wrote:

the c code as follows:

#include <stdio.h>
#include <stdlib.h>

int main()
{
printf("%s\n", getenv("NAME"));
return 0;
}

Segmentation Fault

why?

Perhaps "NAME" is not defined in the environment, so getenv() returns
NULL and your attempt to dereference it is illegal. We just had this a
couple of days ago. That you did not see it shows that you violated
usenet etiqutte by posting without following the newsgroup first. You
probably didn't check the FAQ either.

[Barry Schwarz]

On Fri, 9 Apr 2004 23:51:26 GMT, J Wang <cs****@bath.ac.uk> wrote:

the c code as follows:

#include <stdio.h>
#include <stdlib.h>

int main()
{
printf("%s\n", getenv("NAME"));
return 0;
}

Segmentation Fault

why?

getenv is allowed to return NULL. You need to check for that
situation before dereferencing the return value.
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