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[SUMMARY] Bytecode Compiler (#100)
I'm very glad this quiz came along, because I honestly had no idea what
"bytecode" meant before I read this quiz. I think it was good for dumbing the idea down for the masses (or just me, if everyone else already knew this stuff). Not all of us port entire virtual machines, as the quiz creator does. There were a couple of approaches to this week's quiz. One strategy was to convert the expression into valid Ruby that would emit the bytecode and let Ruby do the parsing. This works for this parsing problem because Ruby's expression rules are comparable to those specified by the quiz. I think we can break down one of those solutions to better understand what was involved with this problem. Here's a chunk of code from Cameron Pope's solution: module CreateExpressions def +(other) Expr.new(:add, self, other) end def -(other) Expr.new(:sub, self, other) end def *(other) Expr.new(:mul, self, other) end def /(other) Expr.new(:div, self, other) end def %(other) Expr.new(:mod, self, other) end def **(other) Expr.new(:pow, self, other) end end You can see a set of operator definitions here, matching those needed by the quiz. Instead of these methods doing what they normally do (math), the versions we see here just create and return an Expr object. Let's have a look at that definition now: class Expr include CreateExpressions OPCODES = {:add => 0x0a, :sub => 0x0b, :mul => 0x0c, :pow => 0x0d, :div => 0x0e, :mod => 0x0f} def initialize(op, a, b) @op = op @first = a @second = b end def to_s "(#{@op.to_s} #{@first.to_s} #{@second.to_s})" end def emit @first.emit << @second.emit << OPCODES[@op] end end Ignoring the to_s() method, which is just for debugging, this object is trivial. It stores an operator and operands, and can emit() bytecode. To do that, it forwards the emit() call to both operands, which may be nested Expr objects (say from parenthetical expressions) or something else we will discuss in just a moment. emit() also looks up the operator's bytecode and appends that to the results of the operand emit()s. Note that this class include()s all of the operator definitions we saw earlier. This means you can add, multiply, etc. Expr objects, which yields a new Expr with the original Epxrs nested in the operands. That covers expressions, but what about plain old numbers? For that, there is another class: class Const include CreateExpressions OPCODES = {2 => 0x01, 4 => 0x02} def initialize(i) @value = i end def to_s @value end def emit case @value when (-32768..32767): bytes = [@value].pack("n").unpack("C*") else bytes = [@value].pack("N").unpack("C*") end bytes.insert 0, OPCODES[bytes.size] end end We have pretty much the same pattern here, save that emit() converts the number to the bytecode for the properly sized constant plus the packed bytes. Again we see the arithmetic operators include()ed, so these too can be combined with other Const and Expr objects. Now we need a means to turn the normal numbers of the expression into Const objects: class Fixnum def to_const Const.new(self) end end Yep, that will do just that. All that's left is to put it to use: class Compiler def self.compile(expr) self.mangle(expr).emit.flatten end def self.explain(expr) self.mangle(expr).to_s end private def self.mangle(expr) eval(expr.gsub(/\d+/) {|s| "#{s}.to_const()"}) end end The mangle() method is the heart of this solution. A simple Regexp is used to tack a to_const() call on to each number of the expression and a hand-off is made to eval() to build up the indicated combination of Const and Expr objects. Once mangle()d, the result is emit()ted and flatten()ed into the final bytecode Array in compile(). That's easy enough to understand and it works just fine in this instance. However, what if the rules differed from Ruby's and you couldn't lean on Ruby's parser? In that case, you would have to roll your own parser, which some decided to do. Luckily there is already a popular algorithm for unrolling infix arithmetic expressions into the RPN order required by bytecode spec: http://en.wikipedia.org/wiki/Shunting_yard_algorithm Several people employed this strategy, including Daniel Martin: # This is a solution to Ruby Quiz #100 # # It's basically just a shunting algorithm, but with a twist # since it needs to distinguish between a "-" that's part of # a number and a "-" that's an operator. To do that, I use # a state machine while parsing to remember if I need next # an operator or an integer. require 'strscan' class Compiler # A small class made so that I can use case ... when # with a StringScanner class Token < Regexp def initialize(re) super(re) end # Using is_a? instead of respond_to? isn't very duck-typey, # but unfortunately String#scan and StringScanner#scan mean # completely different things. def ===(s) if (s.is_a?(StringScanner)) s.scan(self) else super(s) end end end # ... This first class of Daniel's solution is really just a Regexp with a custom case equals method. This sets up an elegant syntax for the solution proper we will encounter in a bit. # ... # The tokens I need WSPACE = Token.new(/\s+/) LPAREN = Token.new(/\(/) RPAREN = Token.new(/\)/) OP = Token.new(/\*\*|[+*%\/-]/) NEG = Token.new(/-/) INT = Token.new(/\d+/) OpValMap = {'+' => 0x0a, '-' => 0x0b, '*' => 0x0c, '**' => 0x0d, '/' => 0x0e, '%' => 0x0f} # ... Here you see the class used to build the Tokens for lexxing the expression content. These are very basic regular expressions. Here are the interface methods required by the quiz solution: # ... def initialize(instring) @scanner = StringScanner.new(instring) @opstack = Array.new @outarr = Array.new end def compile() state = :state_int while state != :state_end case @scanner when WSPACE next else state = send(state) raise "Syntax error at index #{@scanner.pos}" if ! state end end while ! @opstack.empty? op = @opstack.pop raise "Mismatched parens" if LPAREN === op @outarr << OpValMap[op] end @outarr end # Class method as required by the test harness def self.compile(instring) new(instring).compile end # ... Notice that initialize() sets up the stack and output Arrays needed by the Shunting Yard algorithm. The compile() method wraps a state machine we will examine the workings of shortly. You can see that is discards whitespace as needed, forwards to state methods, pop()s the final operators as the algorithm requires, and returns the resulting output Array. The class compile() is just a wrapper over the other two methods. The heart of what is going on here is hidden in the state methods. From compile() we saw that the initial state for the machine is :state_int, so let's begin with that method: # ... private # ... # state where we're expecting an integer or left paren def state_int case @scanner when LPAREN @opstack << @scanner.matched :state_int when INT integer(@scanner.matched.to_i) :state_op when NEG :state_neg end end # ... Here we begin to see the magic of the Token class. Matching a Token advances the StringScanner index and matched() can then be used to grab the element. The LPAREN when clause is right out of the algorithm description, and the INT clause is pretty obviously if you know that integer() handles the constant conversion. The NEG clause is needed to distinguish an unary minus from the binary operator. Note that each case returns the next expected state for the machine. Here's the integer() helper we have seen before: # ... # Handle an integer def integer(i) if (i <= 32767 and i >= -32768) @outarr << 0x01 @outarr.push(*([i].pack("n").unpack("C*"))) else @outarr << 0x02 @outarr.push(*([i].pack("N").unpack("C*"))) end end # ... The only difference here is that constants are pushed onto the output Array as required by the algorithm. Let's return to the state methods: # ... # Expecting an operator or right paren def state_op case @scanner when RPAREN while not LPAREN === @opstack[-1] raise "Mismatched parens" if @opstack.empty? @outarr << OpValMap[@opstack.pop] end @opstack.pop :state_op when OP op = @scanner.matched while is_lower(@opstack[-1], op) @outarr << OpValMap[@opstack.pop] end @opstack << op :state_int else # I would handle this with an EOS token, but unfortunately # StringScanner is broken w.r.t. @scanner.scan(/$/) :state_end if @scanner.eos? end end # ... Again, the RPAREN clause is right out of the algorithm description. The OP clause is as well and you can see that it handles the precedence check via the is_lower() helper method. The else clause gives us our exit state when the expression has been exhausted. Here's the is_lower() helper: # ... # Define the precedence order # One thing to note is that for an operator a, # is_lower(a,a) being true will make that operator # left-associative, while is_lower(a,a) being false # makes that operator right-associative. Note that # we want ** to be right associative, but all other # operators to be left associative. def is_lower(op_on_stack, op_in_hand) case op_on_stack when nil, LPAREN; false when /\*\*|[*\/%]/; op_in_hand =~ /^.$/ when /[+-]/; op_in_hand =~ /[+-]/ end end # ... The comment surely explains this better than I can, but the point of this method is to resolve whether or not the operator we just matched is lower in precedence than the operator on the stack. For example, in the last line, when we have a plus or minus on the stack only another plus or minus would trigger the true result. Here's the final state: # ... # The state where we've seen a minus and are expecting # the rest of the integer def state_neg case @scanner when INT integer(-(@scanner.matched.to_i)) :state_op end end # ... end This just reads the constant following a negation operator and ensures that it is negated. Those are two different approaches that pass the quiz tests. It's probably slightly easier to lean on Ruby in cases like this where you know you can get away with it. If you can't though, the custom parser isn't too much more work as Daniel shows. My thanks to all who taught me many things I didn't know about bytecode compilation through their solutions and to Ross Bamford for the educational quiz. Tomorrow we will play with VCRs and I'm told that dates me... |
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