Calculating the maximum timeout that a user can set via the
SG_SET_TIMEOUT ioctl involves multiplying INT_MAX by USER_HZ/HZ. If
USER_HZ is larger than HZ then this results in an overflow when
performed as a 32 bit integer calculation, resulting in compiler
warnings such as the following:
drivers/scsi/sg.c: In function 'sg_ioctl':
drivers/scsi/sg.c:91:67: warning: integer overflow in expression [-Woverflow]
#define MULDIV(X,MUL,DIV) ((((X % DIV) * MUL) / DIV) + ((X / DIV) * MUL))
^
drivers/scsi/sg.c:887:14: note: in expansion of macro 'MULDIV'
if (val >= MULDIV (INT_MAX, USER_HZ, HZ))
^
drivers/scsi/sg.c:91:67: warning: integer overflow in expression [-Woverflow]
#define MULDIV(X,MUL,DIV) ((((X % DIV) * MUL) / DIV) + ((X / DIV) * MUL))
^
drivers/scsi/sg.c:888:13: note: in expansion of macro 'MULDIV'
val = MULDIV (INT_MAX, USER_HZ, HZ);
^
Avoid this overflow by performing the (constant) arithmetic on 64 bit
integers, which ensures that overflow from multiplying the 32 bit values
cannot occur. When converting the result back to a 32 bit integer use
min_t to ensure that we don't simply truncate a value beyond INT_MAX to
a 32 bit integer, but instead use INT_MAX where the result was larger
than it. As the values are all compile time constant the 64 bit
arithmetic should have no runtime cost.
Signed-off-by: Paul Burton <paul.burton@imgtec.com>
Acked-by: Douglas Gilbert <dgilbert@interlog.com>
Signed-off-by: Martin K. Petersen <martin.petersen@oracle.com>
b9b6e80ad3