Обсуждение: Struggling with c functions
Hi all,
I'm rewriting my OLD crypt (Thanks to Henrique) C_fonction to version 0
forms :
I have this C file compiled OK to a shared library:
/*
*
* Henrique Pantarotto (scanner@cepa.com.br)
* Funcao para encriptar senhas (Function to encrypt passwords)
* September 1999
*
* PS: Note that all crypted passwords are created with salt "HP" (my name
* initials..) You can change that, or if you know C, you can do in a way
* that it will pick two random characters (the way it should really be).
*
*/
#include <strings.h>
#include <unistd.h>
#include <postgres.h>
text *post_crypt(text *user)
{text *password;char * crypt();long now=time((long *) 0);int len;char salt[7]="PY", *crypted;/*strcpy(salt,l64a(now));
salt[3]='\0';*/crypted=crypt(VARDATA(user),salt);len=strlen(crypted);password= palloc((int32) 13 +
VARHDRSZ);VARATT_SIZEP(password)=(int32) VARHDRSZ + 13;memcpy(VARDATA(password),crypted,len);return password;
}
text *sql_crypt(text *user,text *salt)
{ text *password; char * crypt(), *crypted; int len; char s[3]; strncpy(s,VARDATA(salt),2); s[2]='\0';
crypted=crypt(VARDATA(user),s);len=strlen(crypted); password=palloc((int32) 13 + VARHDRSZ);
VARATT_SIZEP(password)=(int32)13 + VARHDRSZ; memcpy(VARDATA(password),crypted,len); return password;
}
/*
Compile using something like this:
gcc -I/home/postgres/postgresql-6.5.1/src/include -I/home/postgres/postgresql-6.5.1/src/backend -O2 -Wall
-Wmissing-prototypes-fpic -I/home/postgres/postgresql-6.5.1/src/include -c -o encrypt.o encrypt.c
gcc -shared -o encrypt.so encrypt.o
And last, you create the trigger in PostgreSQL using this:
create function encrypt(text)
returns text as '/usr/local/pgsql/lib/encrypt.so' language 'c';
If everything is okay, you'll probably have: select encrypt('secret') working
and showing:
encrypt
------------
HPK1Jt2NX21G.
(1 row)
*/
I have defined to SQL function:
CREATE FUNCTION post_crypt(text) RETURNS text AS 'xxxx/encrypt.so'
CREATE FUNCTION sql_cypt(text,text) RETURNS text AS 'xxxx/encrypt.so';
WHY on earth does
SELECT post_crypt('test'),sql_crypt('test','PY')
NOT GIVE the same result???
Please help,
This is most urgent (My customer can't use this function anymore); it
worked OK with 7.0.3!!
Regards,
--
Olivier PRENANT Tel: +33-5-61-50-97-00 (Work)
Quartier d'Harraud Turrou +33-5-61-50-97-01 (Fax)
31190 AUTERIVE +33-6-07-63-80-64 (GSM)
FRANCE Email: ohp@pyrenet.fr
------------------------------------------------------------------------------
Make your life a dream, make your dream a reality. (St Exupery)
You actually almost have it right.
You are passing VARDATA(user) to crypt, this is wrong.
You must do something like this:
int ulen = VARSIZE(user)-VARHDRSZ;
char utmp[ulen+]; // This works in newer GCC, cool.
memcpy(utmp,VARDATA(user), len);
utmp[ulen]=0;
crypted=crypt(utmp,salt);
Strings are not gurenteed to be NULL teminated.
Olivier PRENANT wrote:
>
> Hi all,
>
> I'm rewriting my OLD crypt (Thanks to Henrique) C_fonction to version 0
> forms :
>
> I have this C file compiled OK to a shared library:
>
> /*
> *
> * Henrique Pantarotto (scanner@cepa.com.br)
> * Funcao para encriptar senhas (Function to encrypt passwords)
> * September 1999
> *
> * PS: Note that all crypted passwords are created with salt "HP" (my name
> * initials..) You can change that, or if you know C, you can do in a way
> * that it will pick two random characters (the way it should really be).
> *
> */
>
> #include <strings.h>
> #include <unistd.h>
>
> #include <postgres.h>
>
> text *post_crypt(text *user)
> {
> text *password;
> char * crypt();
> long now=time((long *) 0);
> int len;
> char salt[7]="PY", *crypted;
> /*strcpy(salt,l64a(now));
> salt[3]='\0'; */
> crypted=crypt(VARDATA(user),salt);
> len=strlen(crypted);
> password= palloc((int32) 13 + VARHDRSZ);
> VARATT_SIZEP(password)= (int32) VARHDRSZ + 13;
> memcpy(VARDATA(password),crypted,len);
> return password;
> }
>
> text *sql_crypt(text *user,text *salt)
> {
> text *password;
> char * crypt(), *crypted;
> int len;
> char s[3];
> strncpy(s,VARDATA(salt),2);
> s[2]='\0';
> crypted=crypt(VARDATA(user),s);
> len=strlen(crypted);
> password=palloc((int32) 13 + VARHDRSZ);
> VARATT_SIZEP(password)=(int32) 13 + VARHDRSZ;
> memcpy(VARDATA(password),crypted,len);
> return password;
> }
>
> /*
> Compile using something like this:
>
> gcc -I/home/postgres/postgresql-6.5.1/src/include -I/home/postgres/postgresql-6.5.1/src/backend -O2 -Wall
-Wmissing-prototypes-fpic -I/home/postgres/postgresql-6.5.1/src/include -c -o encrypt.o encrypt.c
> gcc -shared -o encrypt.so encrypt.o
>
> And last, you create the trigger in PostgreSQL using this:
>
> create function encrypt(text)
> returns text as '/usr/local/pgsql/lib/encrypt.so' language 'c';
>
> If everything is okay, you'll probably have: select encrypt('secret') working
> and showing:
>
> encrypt
> ------------
> HPK1Jt2NX21G.
> (1 row)
> */
>
> I have defined to SQL function:
>
> CREATE FUNCTION post_crypt(text) RETURNS text AS 'xxxx/encrypt.so'
> CREATE FUNCTION sql_cypt(text,text) RETURNS text AS 'xxxx/encrypt.so';
>
> WHY on earth does
>
> SELECT post_crypt('test'),sql_crypt('test','PY')
> NOT GIVE the same result???
>
> Please help,
>
> This is most urgent (My customer can't use this function anymore); it
> worked OK with 7.0.3!!
>
> Regards,
> --
> Olivier PRENANT Tel: +33-5-61-50-97-00 (Work)
> Quartier d'Harraud Turrou +33-5-61-50-97-01 (Fax)
> 31190 AUTERIVE +33-6-07-63-80-64 (GSM)
> FRANCE Email: ohp@pyrenet.fr
> ------------------------------------------------------------------------------
> Make your life a dream, make your dream a reality. (St Exupery)
>
> ---------------------------(end of broadcast)---------------------------
> TIP 6: Have you searched our list archives?
>
> http://www.postgresql.org/search.mpl
--
I'm not offering myself as an example; every life evolves by its own laws.
------------------------
http://www.mohawksoft.com
Many thanks to you!!!
It now works (did'nt realize that strings where not null
terminated) stupid me!!!
Regards,
On Sat, 28 Apr 2001, mlw wrote:
> You actually almost have it right.
>
> You are passing VARDATA(user) to crypt, this is wrong.
>
> You must do something like this:
>
> int ulen = VARSIZE(user)-VARHDRSZ;
> char utmp[ulen+]; // This works in newer GCC, cool.
> memcpy(utmp,VARDATA(user), len);
> utmp[ulen]=0;
> crypted=crypt(utmp,salt);
>
> Strings are not gurenteed to be NULL teminated.
>
>
> Olivier PRENANT wrote:
> >
> > Hi all,
> >
> > I'm rewriting my OLD crypt (Thanks to Henrique) C_fonction to version 0
> > forms :
> >
> > I have this C file compiled OK to a shared library:
> >
> > /*
> > *
> > * Henrique Pantarotto (scanner@cepa.com.br)
> > * Funcao para encriptar senhas (Function to encrypt passwords)
> > * September 1999
> > *
> > * PS: Note that all crypted passwords are created with salt "HP" (my name
> > * initials..) You can change that, or if you know C, you can do in a way
> > * that it will pick two random characters (the way it should really be).
> > *
> > */
> >
> > #include <strings.h>
> > #include <unistd.h>
> >
> > #include <postgres.h>
> >
> > text *post_crypt(text *user)
> > {
> > text *password;
> > char * crypt();
> > long now=time((long *) 0);
> > int len;
> > char salt[7]="PY", *crypted;
> > /*strcpy(salt,l64a(now));
> > salt[3]='\0'; */
> > crypted=crypt(VARDATA(user),salt);
> > len=strlen(crypted);
> > password= palloc((int32) 13 + VARHDRSZ);
> > VARATT_SIZEP(password)= (int32) VARHDRSZ + 13;
> > memcpy(VARDATA(password),crypted,len);
> > return password;
> > }
> >
> > text *sql_crypt(text *user,text *salt)
> > {
> > text *password;
> > char * crypt(), *crypted;
> > int len;
> > char s[3];
> > strncpy(s,VARDATA(salt),2);
> > s[2]='\0';
> > crypted=crypt(VARDATA(user),s);
> > len=strlen(crypted);
> > password=palloc((int32) 13 + VARHDRSZ);
> > VARATT_SIZEP(password)=(int32) 13 + VARHDRSZ;
> > memcpy(VARDATA(password),crypted,len);
> > return password;
> > }
> >
> > /*
> > Compile using something like this:
> >
> > gcc -I/home/postgres/postgresql-6.5.1/src/include -I/home/postgres/postgresql-6.5.1/src/backend -O2 -Wall
-Wmissing-prototypes-fpic -I/home/postgres/postgresql-6.5.1/src/include -c -o encrypt.o encrypt.c
> > gcc -shared -o encrypt.so encrypt.o
> >
> > And last, you create the trigger in PostgreSQL using this:
> >
> > create function encrypt(text)
> > returns text as '/usr/local/pgsql/lib/encrypt.so' language 'c';
> >
> > If everything is okay, you'll probably have: select encrypt('secret') working
> > and showing:
> >
> > encrypt
> > ------------
> > HPK1Jt2NX21G.
> > (1 row)
> > */
> >
> > I have defined to SQL function:
> >
> > CREATE FUNCTION post_crypt(text) RETURNS text AS 'xxxx/encrypt.so'
> > CREATE FUNCTION sql_cypt(text,text) RETURNS text AS 'xxxx/encrypt.so';
> >
> > WHY on earth does
> >
> > SELECT post_crypt('test'),sql_crypt('test','PY')
> > NOT GIVE the same result???
> >
> > Please help,
> >
> > This is most urgent (My customer can't use this function anymore); it
> > worked OK with 7.0.3!!
> >
> > Regards,
> > --
> > Olivier PRENANT Tel: +33-5-61-50-97-00 (Work)
> > Quartier d'Harraud Turrou +33-5-61-50-97-01 (Fax)
> > 31190 AUTERIVE +33-6-07-63-80-64 (GSM)
> > FRANCE Email: ohp@pyrenet.fr
> > ------------------------------------------------------------------------------
> > Make your life a dream, make your dream a reality. (St Exupery)
> >
> > ---------------------------(end of broadcast)---------------------------
> > TIP 6: Have you searched our list archives?
> >
> > http://www.postgresql.org/search.mpl
>
>
--
Olivier PRENANT Tel: +33-5-61-50-97-00 (Work)
Quartier d'Harraud Turrou +33-5-61-50-97-01 (Fax)
31190 AUTERIVE +33-6-07-63-80-64 (GSM)
FRANCE Email: ohp@pyrenet.fr
------------------------------------------------------------------------------
Make your life a dream, make your dream a reality. (St Exupery)