strcpy HELP!!!!
I'm using libc6 2.0.7pre1-4 and strcpy gives me a segmentation
error!
I've attached my C test program for inspection(source only).
I only do a few things. malloc memory for a variable(forgot a
corresponding free)
copy a string in this varible, copy this string to another
string, print both
on the screen. The exact error message is attached to the
source.
I will now check bug track.
Thanks in advance
Philip
--
_
_ \ \ From Witomatic Keep,
/ \_\ | Home of Witwerg the Sage
| //o_/ | --
| \ \_/ Weilding GCC and Linux 2.0.33
\ \_Tschuss! Auf Wiedsehen!#include stdlib.h
#include stdio.h
#include string.h
void main(void)
{
char *temp1, *temp2;
temp1 = malloc (10);
strcpy (temp1, high all);
strcpy (temp2, temp1);
printf(%s %s, temp1, temp2);
}
/* error --- program recieved signal SIGSEGV, Segmentation Fault.
* 0x40050be7*/
Re: strcpy HELP!!!!
This is your fault. You need to allocate memory for temp2.
void main(void)
{
char *temp1, *temp2;
temp1 = malloc (10);
insert `temp2 = malloc (10);' here.
strcpy (temp1, high all);
strcpy (temp2, temp1);
printf(%s %s, temp1, temp2);
}
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Re: strcpy HELP!!!!
#include stdlib.h
#include stdio.h
#include string.h
void main(void)
{
char *temp1, *temp2;
temp1 = malloc (10);
strcpy (temp1, high all);
strcpy (temp2, temp1);
printf(%s %s, temp1, temp2);
}
/* error --- program recieved signal SIGSEGV, Segmentation Fault.
* 0x40050be7*/
No wonder. You have to allocate memory for temp2 string before copying to it.
do temp2=malloc(strlen(temp1)+1); in between of two strcpy and everything's
gonna be OK.
Good luck.
Alex Y.
_
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( (o___ +---+
| _ 7 |Alexander Yukhimets|
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RE: strcpy HELP!!!!
Phil, You are copying the string to temp2 which has not been initialized and is a NULL pointer. You need to allocate memory of at least the same size as temp1 to temp2 and this will solve your problem. Scott D. Killen Scott Killen Software http://www.skillsoft.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Phil Sent: Wednesday, April 15, 1998 9:01 PM To: debian-user@lists.debian.org Subject: strcpy HELP I'm using libc6 2.0.7pre1-4 and strcpy gives me a segmentation error! I've attached my C test program for inspection(source only). I only do a few things. malloc memory for a variable(forgot a corresponding free) copy a string in this varible, copy this string to another string, print both on the screen. The exact error message is attached to the source. I will now check bug track. Thanks in advance Philip -- _ _ \ \ From Witomatic Keep, / \_\ | Home of Witwerg the Sage | //o_/ | -- | \ \_/ Weilding GCC and Linux 2.0.33 \ \_Tschuss! Auf Wiedsehen! -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
RE: strcpy HELP!!!
Scott D. Killen writes: You are copying the string to temp2 which has not been initialized and is a NULL pointer. You need to allocate memory of at least the same size as temp1 to temp2 and this will solve your problem. When using C library string functions always allocate memory of at least the length of the string pointed to by temp1 plus 1 byte to hold the string termination character '\0'. As in: temp2 = (char*)malloc(strlen(temp1) + 1); -- =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Lee Brinton | [EMAIL PROTECTED] PGP public key @ http://www.icubed.com/~6xtippet/pgp-public-key.asc --- The objective is not to die for your country, but to make the enemy die for his country. -- Gen. George Patton =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
RE: strcpy HELP!!!
-BEGIN PGP SIGNED MESSAGE- You are correct. I was making specific assumptions based on the code snippet shown. - -Original Message- From: Lee Brinton [mailto:[EMAIL PROTECTED] Sent: Friday, April 17, 1998 12:40 AM To: debian-user Mailing List Subject: RE: strcpy HELP!!! Scott D. Killen writes: You are copying the string to temp2 which has not been initialized and is a NULL pointer. You need to allocate memory of at least the same size as temp1 to temp2 and this will solve your problem. When using C library string functions always allocate memory of at least the length of the string pointed to by temp1 plus 1 byte to hold the string termination character '\0'. As in: temp2 = (char*)malloc(strlen(temp1) + 1); - -- =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Lee Brinton | [EMAIL PROTECTED] PGP public key @ http://www.icubed.com/~6xtippet/pgp-public-key.asc - --- The objective is not to die for your country, but to make the enemy die for his country. -- Gen. George Patton =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED] -BEGIN PGP SIGNATURE- Version: PGP for Personal Privacy 5.0 Charset: noconv iQCVAwUBNTbulgrrqzzf5TWxAQEi2QP/Ye3hKW8cutHTZP415zAmxhBTdElHzJjx e3j0Aq0MJ+gtDFcCNpraNygmoza2kc8s/wFT3k58npNELO63CaPmiFL2HJcQDiqh BAdXHxN9toPVVrCYNNbRBK8b2gIb05nBOk6JsFzhW+uqtBv/Ha3mBT7kN/ae8/u8 w8werd/ml58= =rj5a -END PGP SIGNATURE- -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
Re: strcpy HELP!!!
Lee Brinton: Scott D. Killen writes: You are copying the string to temp2 which has not been initialized and is a NULL pointer. You need to allocate memory of at least the same size as temp1 to temp2 and this will solve your problem. When using C library string functions always allocate memory of at least the length of the string pointed to by temp1 plus 1 byte to hold the string termination character '\0'. As in: temp2 = (char*)malloc(strlen(temp1) + 1); No they don't always. From the strcpy man page: The strcpy() function copies the string pointed to be src (including the terminating `\0' character) to the array pointed to by dest. The strings may not overlap, and the destination string dest must be large enough to receive the copy. You have to allocate the space by yourself. However, there is strdup: NAME strdup - duplicate a string SYNOPSIS #include string.h char *strdup(const char *s); DESCRIPTION The strdup() function returns a pointer to a new string which is a duplicate of the string s. Memory for the new string is obtained with malloc(3), and can be freed with free(3). This function actually does the memory allocation for you. More information for aspiring C-programmers can be found typing `info libc', provided you have the glibcdoc package installed (at least, that is where the info file is kept on a bo system). Eric Meijer -- E.L. Meijer ([EMAIL PROTECTED]) | tel. office +31 40 2472189 Eindhoven Univ. of Technology | tel. lab. +31 40 2475032 Lab. for Catalysis and Inorg. Chem. (TAK) | tel. fax+31 40 2455054 -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
Re: strcpy HELP!!!
Scott D. Killen writes: You are copying the string to temp2 which has not been initialized and is a NULL pointer. You need to allocate memory of at least the same size as temp1 to temp2 and this will solve your problem. When using C library string functions always allocate memory of at least the length of the string pointed to by temp1 plus 1 byte to hold the string termination character '\0'. As in: temp2 = (char*)malloc(strlen(temp1) + 1); When doing strcpy(temp2,temp1) ??? How could it possibly do that? It cannot change the value of temp2, it is passed by value. Please read man strcpy. Good luck. Alex Y. -- _ _( )_ ( (o___ +---+ | _ 7 |Alexander Yukhimets| \()| http://pages.nyu.edu/~aqy6633/ | / \ \ +---+ -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of unsubscribe. Trouble? Contact [EMAIL PROTECTED]
