Why segmentation fault in this simple code?

[Polar]

Hi!

i'm a newbie in C language and i'm writing my first simple codes.

In one of these, my purpose is to append the ascii value of an interger
(example 101 --> e) at the end of a string to obtain a new (longer)
string. Example:

string: languag
letter: e
string (new value): language

In compiling, i received "Segmentation fault" error but i
can't understand why.

This is the very simple code:

main()
{
char* output="languag";
int i = 101;
char* letter;
letter[0]=(char)i;
letter[1]='\0';
printf("%s\n",letter); //prints "e"
//next line is not executed
strcat(output,letter); //segmentation fault printf("%s\n",output);

}

Thank you :)

[Polar]

On Mon, 09 Aug 2004 11:44:08 +0000, Polar wrote:

main()
{
char* output="languag";
int i = 101;
char* letter;
letter[0]=(char)i;
letter[1]='\0';
printf("%s\n",letter); //prints "e"
//next line is not executed
strcat(output,letter); //segmentation fault printf("%s\n",output);

}

Sorry fo incorrect line-wrap.

This is the code:

main()
{
char* output="languag";
int i = 101;
char* letter;
letter[0]=(char)i;
letter[1]='\0';
printf("%s\n",letter); //prints "e"
//next line is not executed
strcat(output,letter); //segmentation fault
printf("%s\n",output);

}

Thank you :)

[Nick Patavalis]

On 2004-08-09, Polar <po***@pole.po> wrote:

Hi!

In compiling, i received "Segmentation fault" error but i
can't understand why.

Probably not "in compiling" but "in running", but anyway.
main()
{
char* output="languag";
int i = 101;
char* letter; ^^^^^^^

This is a pointer to a character. Currently it points to a random
address inside the computer's "memory". It points to an address that
possibly (and probably) does not "belong" to your "program". Or it can
point to an address where other usefull information is stored, like
the code of your program, or other variables of your program, or
anything.
letter[0]=(char)i;
letter[1]='\0';

Now you try to write something to this random address (that probably
doesn't even belong to you), as well as to the address next to it. A
segfault is imminent!

/npat

[Emmanuel Delahaye]

Polar wrote on 09/08/04 :

main()
{
char* output="languag";
string literals are not guaranteed to be writable. Better to define
them read-only:

char const * output="languag";
int i = 101;
What is this magic value supposed to represent? If you want 'e', just
use 'e'.
char* letter;
Warning. This variable is not initialized, meaning that its value is
undetermined. Being a pointer, it points anywhere.
letter[0]=(char)i;
Dereferencing an undefined pointer invokes an undefined behaviour (UB).
Your program is dead. Fix it.
letter[1]='\0';
UB again.
printf("%s\n",letter); //prints "e"
UB again. (Passing an undefined value to a function)
//next line is not executed
strcat(output,letter); //segmentation fault
UB again for many reasons:

- 'letter' is still undefined
- 'output' is not writable
- Should it be writable, there is no room enough for "e"
printf("%s\n",output);
}

You must use array oc ahrs to do what you want. Memory space doesn't
come by magic.

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

"C is a sharp tool"

[Ravi Uday]

"Polar" <po***@pole.po> wrote in message
news:pa****************************@pole.po...

On Mon, 09 Aug 2004 11:44:08 +0000, Polar wrote:

main()
{
char* output="languag";
int i = 101;
char* letter;
letter[0]=(char)i;
letter[1]='\0';
printf("%s\n",letter); //prints "e"
//next line is not executed
strcat(output,letter); //segmentation fault printf("%s\n",output);

}
Sorry fo incorrect line-wrap.

This is the code:

main()
{
char* output="languag";

'output' is readonly !!
int i = 101;
char* letter;
letter[0]=(char)i;
/* No need of a cast here*/
letter[1]='\0';
printf("%s\n",letter); //prints "e"
//next line is not executed
strcat(output,letter); //segmentation fault
Sure it is segmentation fault..Change the declaration to
char output[32]="languag"; /* Note 32 is a magic number (avoid it
whereever possible */
Ditto for char *letter; change to
char letter[32];
printf("%s\n",output);

}

Thank you :)

[Emmanuel Delahaye]

Ravi Uday wrote on 09/08/04 :

int i = 101;
char* letter;
letter[0]=(char)i;

/* No need of a cast here*/

Yes, but you missed the point. 'letter' is not initialized. The array
is no-existing. 'letter[0]' invokes a UB.

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

"C is a sharp tool"

[Grumble]

Emmanuel Delahaye wrote:

The array is no-existing.

I'd write inexistent (close to French spelling) or nonexistent ;-)

[Emmanuel Delahaye]

Grumble wrote on 09/08/04 :

Emmanuel Delahaye wrote:

The array is no-existing.

I'd write inexistent (close to French spelling) or nonexistent ;-)

Yes, sorry. Thanks for correcting me. I know it's boring to read bad
English.

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

"C is a sharp tool"

[Christopher Benson-Manica]

Polar <po***@pole.po> spoke thus:

main()

main() returns an int. The return type will not default to int under
C99, and it's a good idea to specify its return value even under C89.
Prefer the following:

int main( void )

Other people have already pointed out the actual error(s) in your
code.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

[Emmanuel Delahaye]

(supersedes <mn***********************@YOURBRAnoos.fr>)

Polar wrote on 09/08/04 :

main()
{
char* output="languag";
string literals are not guaranteed to be writable. Better to define
them read-only:

char const * output="languag";
int i = 101;
What is this magic value supposed to represent? If you want 'e', just
use 'e'.
char* letter;
Warning. This variable is not initialized, meaning that its value is
undetermined. Being a pointer, it points anywhere.
letter[0]=(char)i;
Dereferencing an undefined pointer invokes an undefined behaviour (UB).
Your program is dead. Fix it.
letter[1]='\0';
UB again.
printf("%s\n",letter); //prints "e"
UB again. (Passing an undefined value to a function)
//next line is not executed
strcat(output,letter); //segmentation fault
UB again for many reasons:

- 'letter' is still undefined
- 'output' is not writable
- Should it be writable, there is no room enough for "e"
printf("%s\n",output); }

You must use array of chars to do what you want. Memory space doesn't
come by magic.

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

"C is a sharp tool"

[Polar]

[cut]

Thanks to all my teachers :)

Now I understand my errors.

[Polar]

[cut]

Thanks a lot to everyone :) Now I understood my errors :)