util.strbitop: Add common_prefix_bits() method

This returns the number of bits that two strings have in common. It is
significantly more efficient than similar calculations in Lua.

1 files changed, 44 insertions, 0 deletions

diff --git a/spec/util_strbitop_spec.lua b/spec/util_strbitop_spec.lua
index 58a13772..fdb21414 100644
--- a/spec/util_strbitop_spec.lua
+++ b/spec/util_strbitop_spec.lua
@@ -38,4 +38,48 @@ describe("util.strbitop", function ()
 			assert.equal("hello", strbitop.sxor("hello", ""));
 		end);
 	end);
+
+	describe("common_prefix_bits()", function ()
+		local function B(s)
+			assert(#s%8==0, "Invalid test input: B(s): s should be a multiple of 8 bits in length");
+			local byte = 0;
+			local out_str = {};
+			for i = 1, #s do
+				local bit_ascii = s:byte(i);
+				if bit_ascii == 49 then -- '1'
+					byte = byte + 2^((7-(i-1))%8);
+				elseif bit_ascii ~= 48 then
+					error("Invalid test input: B(s): s should contain only '0' or '1' characters");
+				end
+				if (i-1)%8 == 7 then
+					table.insert(out_str, string.char(byte));
+					byte = 0;
+				end
+			end
+			return table.concat(out_str);
+		end
+
+		local _cpb = strbitop.common_prefix_bits;
+		local function test(a, b)
+			local Ba, Bb = B(a), B(b);
+			local ret1 = _cpb(Ba, Bb);
+			local ret2 = _cpb(Bb, Ba);
+			assert(ret1 == ret2, ("parameter order should not make a difference to the result (%s, %s) = %d, reversed = %d"):format(a, b, ret1, ret2));
+			return ret1;
+		end
+		local hex = require "util.hex";
+		it("works on single bytes", function ()
+			assert.equal(0, test("00000000", "11111111"));
+			assert.equal(1, test("10000000", "11111111"));
+			assert.equal(0, test("01000000", "11111111"));
+			assert.equal(0, test("01000000", "11111111"));
+			assert.equal(8, test("11111111", "11111111"));
+		end);
+
+		it("works on multiple bytes", function ()
+			for i = 0, 16 do
+				assert.equal(i, test(string.rep("1", i)..string.rep("0", 16-i), "1111111111111111"));
+			end
+		end);
+	end);
 end);