powercap: intel_rapl: Fix handling for large time window

When setting the power limit time window, software updates the 'y' bits
and 'f' bits in the power limit register, and the value hardware takes
follows the formula below

	Time window = 2 ^ y * (1 + f / 4) * Time_Unit

When handling large time window input from userspace, using left
shifting breaks in two cases:

 1. when ilog2(value) is bigger than 31, in expression "1 << y", left
    shifting by more than 31 bits has undefined behavior. This breaks
    'y'. For example, on an Alderlake platform, "1 << 32" returns 1.

 2. when ilog2(value) equals 31, "1 << 31" returns negative value
    because '1' is recognized as signed int. And this breaks 'f'.

Given that 'y' has 5 bits and hardware can never take a value larger
than 31, fix the first problem by clamp the time window to the maximum
possible value that the hardware can take.

Fix the second problem by using unsigned bit left shift.

Note that hardware has its own maximum time window limitation, which
may be lower than the time window value retrieved from the power limit
register. When this happens, hardware clamps the input to its maximum
time window limitation. That is why a software clamp is preferred to
handle the problem on hand.
Signed-off-by: Zhang Rui <rui.zhang@intel.com>
[ rjw: Adjusted the comment added by this change ]
Signed-off-by: Rafael J. Wysocki <rafael.j.wysocki@intel.com>

drivers/powercap/intel_rapl_common.c

@@ -999,7 +999,15 @@ static u64 rapl_compute_time_window_core(struct rapl_package *rp, u64 value,
 
 		do_div(value, rp->time_unit);
 		y = ilog2(value);
-		f = div64_u64(4 * (value - (1 << y)), 1 << y);
+
+		/*
+		 * The target hardware field is 7 bits wide, so return all ones
+		 * if the exponent is too large.
+		 */
+		if (y > 0x1f)
+			return 0x7f;
+
+		f = div64_u64(4 * (value - (1ULL << y)), 1ULL << y);
 		value = (y & 0x1f) | ((f & 0x3) << 5);
 	}
 	return value;