vendored parslet, deemed stable enough and better without dependency

lib/parslet/source/line_cache.rb

@@ -0,0 +1,96 @@
+
+
+class Parslet::Source
+  # A cache for line start positions. 
+  #
+  class LineCache 
+    def initialize
+      # Stores line endings as a simple position number. The first line always
+      # starts at 0; numbers beyond the biggest entry are on any line > size, 
+      # but probably make a scan to that position neccessary.
+      @line_ends = []
+      @line_ends.extend RangeSearch
+    end
+
+    # Returns a <line, column> tuple for the given input position. 
+    # 
+    def line_and_column(pos)
+      eol_idx = @line_ends.lbound(pos)
+
+      if eol_idx
+        # eol_idx points to the offset that ends the current line.
+        # Let's try to find the offset that starts it: 
+        offset = eol_idx>0 && @line_ends[eol_idx-1] || 0
+        return [eol_idx+1, pos-offset+1]
+      else
+        # eol_idx is nil, that means that we're beyond the last line end that
+        # we know about. Pretend for now that we're just on the last line.
+        offset = @line_ends.last || 0
+        return [@line_ends.size+1, pos-offset+1]
+      end
+    end
+
+    def scan_for_line_endings(start_pos, buf)
+      return unless buf
+
+      buf = StringScanner.new(buf)
+      return unless buf.exist?(/\n/)
+
+      ## If we have already read part or all of buf, we already know about
+      ## line ends in that portion. remove it and correct cur (search index)
+      if @last_line_end && start_pos < @last_line_end
+        # Let's not search the range from start_pos to last_line_end again.
+        buf.pos = @last_line_end - start_pos
+      end
+
+      ## Scan the string for line endings; store the positions of all endings
+      ## in @line_ends. 
+      while buf.skip_until(/\n/)
+        @last_line_end = start_pos + buf.pos
+        @line_ends << @last_line_end
+      end
+    end
+  end
+
+  # Mixin for arrays that implicitly give a number of ranges, where one range
+  # begins where the other one ends.
+  # 
+  #   Example: 
+  #
+  #     [10, 20, 30]
+  #     # would describe [0, 10], (10, 20], (20, 30]
+  #
+  module RangeSearch 
+    def find_mid(left, right)
+      # NOTE: Jonathan Hinkle reported that when mathn is required, just
+      # dividing and relying on the integer truncation is not enough.
+      left + ((right - left) / 2).floor
+    end  
+    
+    # Scans the array for the first number that is > than bound. Returns the 
+    # index of that number. 
+    #
+    def lbound(bound)
+      return nil if empty?
+      return nil unless last > bound
+
+      left = 0
+      right = size - 1 
+
+      loop do
+        mid = find_mid(left, right)
+
+        if self[mid] > bound
+          right = mid
+        else
+          # assert: self[mid] <= bound
+          left = mid+1
+        end
+
+        if right <= left
+          return right
+        end
+      end
+    end
+  end
+end