LKM programming - problem with printk();

08-07-2004	#1 (permalink)
apacz Just Joined! Join Date: Aug 2004 Posts: 5	LKM programming - problem with printk(); Hello, I've got a problem with printk() function. Im trying to print some buffor char by char like: Code: for(i=0;i<=len;i++) printk(KERN_ALERT "%c", k_space[i]); But then i see in /var/log/syslog something like: Aug 7 11:52:01 dstool kernel: i<1>n<1>s<1>m<1>o<1>d<1><1><1> Aug 7 11:52:01 dstool kernel: e<1>.<1>o<1><1><1> With some extra "<1>". Why it appears? When i print k_space as one string i don't have this "<1>". At first i though it was some problem with user space <-> kernel space but it transformed orginall buffor to k_space like: Code: if ((k_space=(char )kmalloc(len, GFP_KERNEL))==NULL) return -1; memcpy_fromfs((void )k_space, (void *)argv[argc], len); And it didnt solve the problem. Any ideas what to do with this ?

08-07-2004	#2 (permalink)
scientica Linux Enthusiast Join Date: Sep 2003 Location: South- or "Mid-" Sweden Posts: 742	I think printk maybe atdds some message count. And as you're printing the message char by char I think that's why you get it after each char. Isn't k_space a 0-terminated string? if so somply do: Code: printk(KERN_ALERT "%s", k_space); or you could do this: Code: char tem_char; temp_char=k_space[len]; // maybe you could do botthia and above on one line. k_space[len]='\0'; // terminate the string printk(KERN_ALERT "%s", k_space); k_space[len]=temp_char; // restore the char we replaced with a 0-terminator either way try to unroll the loop - it's faster. (loops contains jumps, and jumps are 'realativle slow' - speaking about lowlevel stuff as branch prediction, the cosst of a jump (predicted or misspredited, etc) ). __________________ Regards Scienitca (registered user #335819 - http://counter.li.org ) -- A master is nothing more than a student who knows something of which he can teach to other students.

08-07-2004

#3 (permalink)

apacz

Just Joined!

Join Date: Aug 2004

Posts: 5

Hello,

Quote:

Isn't k_space a 0-terminated string? if so somply do:

I wrote that if i prinkt it as a one string (same as you wrote) there is no extra crap like "<1>".
It is important for to have access to simple char. I think "<1>" is connected in some way with this char becouse it doesnt appears only with printk. It doesnt allow me to compare it with other simple chars.
Any idea?