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+// @ts-check
+
+/**
+ * We must remap all the old bits to new bits for each set variant
+ * Only arbitrary variants are considered as those are the only
+ * ones that need to be re-sorted at this time
+ *
+ * An iterated process that removes and sets individual bits simultaneously
+ * will not work because we may have a new bit that is also a later old bit
+ * This means that we would be removing a previously set bit which we don't
+ * want to do
+ *
+ * For example (assume `bN` = `1<<N`)
+ * Given the "total" mapping `[[b1, b3], [b2, b4], [b3, b1], [b4, b2]]`
+ * The mapping is "total" because:
+ * 1. Every input and output is accounted for
+ * 2. All combinations are unique
+ * 3. No one input maps to multiple outputs and vice versa
+ * And, given an offset with all bits set:
+ * V = b1 | b2 | b3 | b4
+ *
+ * Let's explore the issue with removing and setting bits simultaneously:
+ * V & ~b1 | b3 = b2 | b3 | b4
+ * V & ~b2 | b4 = b3 | b4
+ * V & ~b3 | b1 = b1 | b4
+ * V & ~b4 | b2 = b1 | b2
+ *
+ * As you can see, we end up with the wrong result.
+ * This is because we're removing a bit that was previously set.
+ * And, thus the final result is missing b3 and b4.
+ *
+ * Now, let's explore the issue with removing the bits first:
+ * V & ~b1 = b2 | b3 | b4
+ * V & ~b2 = b3 | b4
+ * V & ~b3 = b4
+ * V & ~b4 = 0
+ *
+ * And then setting the bits:
+ * V | b3 = b3
+ * V | b4 = b3 | b4
+ * V | b1 = b1 | b3 | b4
+ * V | b2 = b1 | b2 | b3 | b4
+ *
+ * We get the correct result because we're not removing any bits that were
+ * previously set thus properly remapping the bits to the new order
+ *
+ * To collect this into a single operation that can be done simultaneously
+ * we must first create a mask for the old bits that are set and a mask for
+ * the new bits that are set. Then we can remove the old bits and set the new
+ * bits simultaneously in a "single" operation like so:
+ * OldMask = b1 | b2 | b3 | b4
+ * NewMask = b3 | b4 | b1 | b2
+ *
+ * So this:
+ * V & ~oldMask | newMask
+ *
+ * Expands to this:
+ * V & ~b1 & ~b2 & ~b3 & ~b4 | b3 | b4 | b1 | b2
+ *
+ * Which becomes this:
+ * b1 | b2 | b3 | b4
+ *
+ * Which is the correct result!
+ *
+ * @param {bigint} num
+ * @param {[bigint, bigint][]} mapping
+ */
+export function remapBitfield(num, mapping) {
+  // Create masks for the old and new bits that are set
+  let oldMask = 0n
+  let newMask = 0n
+  for (let [oldBit, newBit] of mapping) {
+    if (num & oldBit) {
+      oldMask = oldMask | oldBit
+      newMask = newMask | newBit
+    }
+  }
+
+  // Remove all old bits
+  // Set all new bits
+  return (num & ~oldMask) | newMask
+}