Proper way to use copy_to_user?

Question

I'm trying to define a system call that modifies the character buffer passed to it. Specifically, something like this:

...
asmlinkage int sys_mycall( char __user *buff, int len )
{
   char tmp[1000];
   copy_from_user(tmp, buff, len);
   /* change tmp here */
   copy_to_user( buff, &tmp, len );
}

Here, copy_to_user returns -1, and the buffer from the calling program is unchanged. What's happening?

The first thing I as a user would do is call the following: char buf[2000]; sys_mycall(buf, 2000); What should happen? — Pawel Marciniak, Dec 02 '09 at 23:58
Can you show the userspace code that demonstrates the problem? — caf, Dec 03 '09 at 04:20
how can it be used to copy to buff[index] of userBuff ? [index != 0] copy_to_user( buff+index, &tmp, len ); — Atheel Massalha, Jan 21 '20 at 07:46

score 8 · Answer 1 · answered Sep 18 '10 at 21:01

8

Remeber that tmp is already a pointer! Correct way to do it:

copy_to_user( buff, tmp, len );

answered Sep 18 '10 at 21:01

Alex

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score 5 · Accepted Answer · answered Dec 02 '09 at 23:39

5

That looks OK. It's possible that the buffer that userspace passed is mapped read-only - for example if it's in the text segment (eg. a string literal). By the way, this is probably what you want:

return copy_to_user(buff, &tmp, len) ? -EFAULT : 0;

answered Dec 02 '09 at 23:39

caf

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The actual code does do that. I simplified the example to illustrate the core issue. – Stefan Kendall Dec 03 '09 at 03:45
3

I abhor C. Your answer fully solves my problem, but I refuse to admit why. – Stefan Kendall Dec 03 '09 at 04:36
1

I think someone forgot to sign their errno :) – Tim Post Dec 03 '09 at 09:04

Proper way to use copy_to_user?

2 Answers2