Confused about usercopy_64.c
I'm confused about something in usercopy_64.c:
unsigned long copy_from_user(void *to, const void __user *from, unsigned long n)
{
if (likely(access_ok(VERIFY_READ, from, n)))
n = __copy_from_user(to, from, n);
else
memset(to, 0, n);
return n;
}If access_ok() returns false, then that means that we cannot copy the data from user-space. So why are we returning 'n'? Shouldn't we return zero, to let the caller know that the function failed?
Signature · 2 lines
Timur Tabi Linux kernel developer at Freescale
Quoted 11 lines · Timur Tabi
I'm confused about something in
usercopy_64.c:unsigned long
copy_from_user(void *to, const void __user *from, unsigned long n){ if (likely(access_ok(VERIFY_READ, from, n))) n = __copy_from_user(to, from, n); else memset(to, 0, n); return n; }
Quoted 3 lines
If
access_ok()returns false, then that means that we cannot copy the data from user-space. So why are we returning 'n'? Shouldn't we return zero, to let the caller know that the function failed?
copy_from_user() returns number of bytes _not_ copied.
-Olof
Quoted 11 lines · Timur Tabi
I'm confused about something in
usercopy_64.c:unsigned long
copy_from_user(void *to, const void __user *from, unsigned long n){ if (likely(access_ok(VERIFY_READ, from, n))) n = __copy_from_user(to, from, n); else memset(to, 0, n); return n; }
Quoted 2 lines
If
access_ok()returns false, then that means that we cannot copy the data from user-space. So why are we returning 'n'?
To tell the caller that the function failed. The result is the number of bytes _left_ to be copying, ie. zero means success.
Andreas.
Signature · 4 lines
Andreas Schwab, SuSE Labs, [email protected] SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany PGP key fingerprint = 58CA 54C7 6D53 942B 1756 01D3 44D5 214B 8276 4ED5 "And now for something completely different."