string_helpers: fix precision loss for some inputs

It was noticed that we lose precision in the final calculation for some
inputs.  The most egregious example is size=3000 blk_size=1900 in units
of 10 should yield 5.70 MB but in fact yields 3.00 MB (oops).

This is because the current algorithm doesn't correctly account for
all the remainders in the logarithms.  Fix this by doing a correct
calculation in the remainders based on napier's algorithm.

Additionally, now we have the correct result, we have to account for
arithmetic rounding because we're printing 3 digits of precision.  This
means that if the fourth digit is five or greater, we have to round up,
so add a section to ensure correct rounding.  Finally account for all
possible inputs correctly, including zero for block size.

Fixes: b9f28d8
Signed-off-by: James Bottomley <[email protected]>
Reported-by: Vitaly Kuznetsov <[email protected]>
Cc: <[email protected]>	[delay until after 4.4 release]
Signed-off-by: Andrew Morton <[email protected]>
Signed-off-by: Linus Torvalds <[email protected]>

lib/string_helpers.c

@@ -43,50 +43,73 @@ void string_get_size(u64 size, u64 blk_size, const enum string_size_units units,
 		[STRING_UNITS_10] = 1000,
 		[STRING_UNITS_2] = 1024,
 	};
-	int i, j;
-	u32 remainder = 0, sf_cap, exp;
+	static const unsigned int rounding[] = { 500, 50, 5 };
+	int i = 0, j;
+	u32 remainder = 0, sf_cap;
 	char tmp[8];
 	const char *unit;
 
 	tmp[0] = '\0';
-	i = 0;
-	if (!size)
+
+	if (blk_size == 0)
+		size = 0;
+	if (size == 0)
 		goto out;
 
-	while (blk_size >= divisor[units]) {
-		remainder = do_div(blk_size, divisor[units]);
+	/* This is Napier's algorithm.  Reduce the original block size to
+	 *
+	 * coefficient * divisor[units]^i
+	 *
+	 * we do the reduction so both coefficients are just under 32 bits so
+	 * that multiplying them together won't overflow 64 bits and we keep
+	 * as much precision as possible in the numbers.
+	 *
+	 * Note: it's safe to throw away the remainders here because all the
+	 * precision is in the coefficients.
+	 */
+	while (blk_size >> 32) {
+		do_div(blk_size, divisor[units]);
 		i++;
 	}
 
-	exp = divisor[units] / (u32)blk_size;
-	/*
-	 * size must be strictly greater than exp here to ensure that remainder
-	 * is greater than divisor[units] coming out of the if below.
-	 */
-	if (size > exp) {
-		remainder = do_div(size, divisor[units]);
-		remainder *= blk_size;
+	while (size >> 32) {
+		do_div(size, divisor[units]);
 		i++;
-	} else {
-		remainder *= size;
 	}
 
+	/* now perform the actual multiplication keeping i as the sum of the
+	 * two logarithms */
 	size *= blk_size;
-	size += remainder / divisor[units];
-	remainder %= divisor[units];
 
+	/* and logarithmically reduce it until it's just under the divisor */
 	while (size >= divisor[units]) {
 		remainder = do_div(size, divisor[units]);
 		i++;
 	}
 
+	/* work out in j how many digits of precision we need from the
+	 * remainder */
 	sf_cap = size;
 	for (j = 0; sf_cap*10 < 1000; j++)
 		sf_cap *= 10;
 
-	if (j) {
+	if (units == STRING_UNITS_2) {
+		/* express the remainder as a decimal.  It's currently the
+		 * numerator of a fraction whose denominator is
+		 * divisor[units], which is 1 << 10 for STRING_UNITS_2 */
 		remainder *= 1000;
-		remainder /= divisor[units];
+		remainder >>= 10;
+	}
+
+	/* add a 5 to the digit below what will be printed to ensure
+	 * an arithmetical round up and carry it through to size */
+	remainder += rounding[j];
+	if (remainder >= 1000) {
+		remainder -= 1000;
+		size += 1;
+	}
+
+	if (j) {
 		snprintf(tmp, sizeof(tmp), ".%03u", remainder);
 		tmp[j+1] = '\0';
 	}