why this crash?

[linq936]

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;
}

DO you see why?

[Harald van =?UTF-8?B?RMSzaw==?=]

li*****@hotmail .com wrote:

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;
}

After removing the "<=====cras h!":

test.c: In function ‘lower’:
test.c:4: warning: implicit declaration of function ‘tolower’
/usr/lib/gcc/x86_64-pc-linux-gnu/4.1.2/../../../../lib64/crt1.o: In function
`_start':
(.text+0x20): undefined reference to `main'
collect2: ld returned 1 exit status

Please, when possible, post complete compilable programs.

DO you see why?

You have not posted enough code to make this clear. One likely possibility
is explained in the FAQ at <http://c-faq.com/decl/strlitinit.html >, but
there are other possibilities. If this is what happened, then the problem
is not in the code that you posted, but in how you call the lower function.
If you had posted a complete compilable program, it would have been
possible to be sure if this is what happened.

[Keith Thompson]

li*****@hotmail .com writes:

It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;
}

DO you see why?

No, because you didn't post the entire program, so we can't see where
the actual error is.

But I note that if you call your lower() function with a string
literal as the argument, a crash is very likely.

You should also be careful with the argument you pass to tolower(). A
negative value other than EOF causes undefined behavior; if plain char
is signed on your system, that could also cause a crash.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[sid]

On Jul 3, 5:37 am, linq...@hotmail .com wrote:

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;

}

DO you see why?

I am not sure but I think the argument is supposed to be passed like
char *str and not char* str. If I am wrong kindly correct me and also
please tell me the difference between the two.
Thank you

[Martin Ambuhl]

sid wrote:

I am not sure but I think the argument is supposed to be passed like
char *str and not char* str. If I am wrong kindly correct me and also
please tell me the difference between the two.

a) You are wrong
b) There is no difference

Thank you

[Robert Gamble]

On Jul 2, 11:01 pm, sid <kingsiddha...@ gmail.comwrote:

On Jul 3, 5:37 am, linq...@hotmail .com wrote:

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;

}

DO you see why?

I am not sure but I think the argument is supposed to be passed like
char *str and not char* str.

You are incorrect.

If I am wrong kindly correct me and also
please tell me the difference between the two.

There is no difference in this context, it is a matter of style.

Robert Gamble

[J. J. Farrell]

On Jul 2, 8:01 pm, sid <kingsiddha...@ gmail.comwrote:

On Jul 3, 5:37 am, linq...@hotmail .com wrote:

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;

}

DO you see why?

I am not sure but I think the argument is supposed to be passed like
char *str and not char* str. If I am wrong kindly correct me

Consider yourself kindly corrected.

and also
please tell me the difference between the two.

The characters '*' and ' ' are transposed - that's the only difference.

[Default User]

li*****@hotmail .com wrote:

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;
}

DO you see why?

No. You only showed us part of the code. Post a complete, minimal
program that demonstrates the problem.

I'll bet you passed it a string literal.

Brian

[Army1987]

<li*****@hotmai l.comha scritto nel messaggio news:11******** **************@ o11g2000prd.goo glegroups.com.. .

Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!

Are plain chars signed on your system?
Try *str = (char)tolower(( unsigned char)*str);

str++;
}
return p;
}

DO you see why?

[Army1987]

"sid" <ki***********@ gmail.comha scritto nel messaggio news:11******** **************@ g37g2000prf.goo glegroups.com.. .

On Jul 3, 5:37 am, linq...@hotmail .com wrote:

>Hi,
It could be a very simple problem, but i can not see it...

Here is the code:

char* lower(char* str){
char* p = str;
while ( *str ) {
*str = tolower(*str); <=====crash!
str++;
}
return p;

}

DO you see why?

I am not sure but I think the argument is supposed to be passed like
char *str and not char* str. If I am wrong kindly correct me and also
please tell me the difference between the two.

None. You could also use char*str. Whitespace isn't significant in C.

Thank you