third_party/python/cpplint/cpplint.py
def FindStartOfExpressionInLine(line, endpos, stack):
  """Find position at the matching start of current expression.

  This is almost the reverse of FindEndOfExpressionInLine, but note
  that the input position and returned position differs by 1.

  Args:
    line: a CleansedLines line.
    endpos: start searching at this position.
    stack: nesting stack at endpos.

  Returns:
    On finding matching start: (index at matching start, None)
    On finding an unclosed expression: (-1, None)
    Otherwise: (-1, new stack at beginning of this line)
  """
  i = endpos
  while i >= 0:
    char = line[i]
    if char in ')]}':
      # Found end of expression, push to expression stack
      stack.append(char)
    elif char == '>':
      # Found potential end of template argument list.
      #
      # Ignore it if it's a "->" or ">=" or "operator>"
      if (i > 0 and
          (line[i - 1] == '-' or
           Match(r'\s>=\s', line[i - 1:]) or
           Search(r'\boperator\s*$', line[0:i]))):
        i -= 1
      else:
        stack.append('>')
    elif char == '<':
      # Found potential start of template argument list
      if i > 0 and line[i - 1] == '<':
        # Left shift operator
        i -= 1
      else:
        # If there is a matching '>', we can pop the expression stack.
        # Otherwise, ignore this '<' since it must be an operator.
        if stack and stack[-1] == '>':
          stack.pop()
          if not stack:
            return (i, None)
    elif char in '([{':
      # Found start of expression.
      #
      # If there are any unmatched '>' on the stack, they must be
      # operators.  Remove those.
      while stack and stack[-1] == '>':
        stack.pop()
      if not stack:
        return (-1, None)
      if ((char == '(' and stack[-1] == ')') or
          (char == '[' and stack[-1] == ']') or
          (char == '{' and stack[-1] == '}')):
        stack.pop()
        if not stack:
          return (i, None)
      else:
        # Mismatched parentheses
        return (-1, None)
    elif char == ';':
      # Found something that look like end of statements.  If we are currently
      # expecting a '<', the matching '>' must have been an operator, since
      # template argument list should not contain statements.
      while stack and stack[-1] == '>':
        stack.pop()
      if not stack:
        return (-1, None)

    i -= 1

  return (-1, stack)
def FindStartOfExpressionInLine(line, endpos, stack):
  """Find position at the matching start of current expression.

  This is almost the reverse of FindEndOfExpressionInLine, but note
  that the input position and returned position differs by 1.

  Args:
    line: a CleansedLines line.
    endpos: start searching at this position.
    stack: nesting stack at endpos.

  Returns:
    On finding matching start: (index at matching start, None)
    On finding an unclosed expression: (-1, None)
    Otherwise: (-1, new stack at beginning of this line)
  """
  i = endpos
  while i >= 0:
    char = line[i]
    if char in ')]}':
      # Found end of expression, push to expression stack
      stack.append(char)
    elif char == '>':
      # Found potential end of template argument list.
      #
      # Ignore it if it's a "->" or ">=" or "operator>"
      if (i > 0 and
          (line[i - 1] == '-' or
           Match(r'\s>=\s', line[i - 1:]) or
           Search(r'\boperator\s*$', line[0:i]))):
        i -= 1
      else:
        stack.append('>')
    elif char == '<':
      # Found potential start of template argument list
      if i > 0 and line[i - 1] == '<':
        # Left shift operator
        i -= 1
      else:
        # If there is a matching '>', we can pop the expression stack.
        # Otherwise, ignore this '<' since it must be an operator.
        if stack and stack[-1] == '>':
          stack.pop()
          if not stack:
            return (i, None)
    elif char in '([{':
      # Found start of expression.
      #
      # If there are any unmatched '>' on the stack, they must be
      # operators.  Remove those.
      while stack and stack[-1] == '>':
        stack.pop()
      if not stack:
        return (-1, None)
      if ((char == '(' and stack[-1] == ')') or
          (char == '[' and stack[-1] == ']') or
          (char == '{' and stack[-1] == '}')):
        stack.pop()
        if not stack:
          return (i, None)
      else:
        # Mismatched parentheses
        return (-1, None)
    elif char == ';':
      # Found something that look like end of statements.  If we are currently
      # expecting a '<', the matching '>' must have been an operator, since
      # template argument list should not contain statements.
      while stack and stack[-1] == '>':
        stack.pop()
      if not stack:
        return (-1, None)

    i -= 1

  return (-1, stack)
Find position at the matching start of current expression.

  This is almost the reverse of FindEndOfExpressionInLine, but note
  that the input position and returned position differs by 1.

  Args:
    line: a CleansedLines line.
    endpos: start searching at this position.
    stack: nesting stack at endpos.

  Returns:
    On finding matching start: (index at matching start, None)
    On finding an unclosed expression: (-1, None)
    Otherwise: (-1, new stack at beginning of this line)