milvus-io / milvus

    return collapsed

def FindEndOfExpressionInLine(line, startpos, stack):

  """Find the position just after the end of current parenthesized expression.

  Args:

    line: a CleansedLines line.

    startpos: start searching at this position.

    stack: nesting stack at startpos.

  Returns:

    On finding matching end: (index just after matching end, None)

    On finding an unclosed expression: (-1, None)

    Otherwise: (-1, new stack at end of this line)

  """

  for i in xrange(startpos, len(line)):

    char = line[i]

    if char in '([{':

      # Found start of parenthesized expression, push to expression stack

      stack.append(char)

    elif char == '<':

      # Found potential start of template argument list

      if i > 0 and line[i - 1] == '<':

        # Left shift operator

        if stack and stack[-1] == '<':

          stack.pop()

          if not stack:

            return (-1, None)

      elif i > 0 and Search(r'\boperator\s*$', line[0:i]):

        # operator<, don't add to stack

        continue

      else:

        # Tentative start of template argument list

        stack.append('<')

    elif char in ')]}':

      # Found end of parenthesized expression.

      #

      # If we are currently expecting a matching '>', the pending '<'

      # must have been an operator.  Remove them from expression stack.

      while stack and stack[-1] == '<':

        stack.pop()

      if not stack:

        return (-1, None)

      if ((stack[-1] == '(' and char == ')') or

          (stack[-1] == '[' and char == ']') or

          (stack[-1] == '{' and char == '}')):

        stack.pop()

        if not stack:

          return (i + 1, None)

      else:

        # Mismatched parentheses

        return (-1, None)

    elif char == '>':

      # Found potential end of template argument list.

      # Ignore "->" and operator functions

      if (i > 0 and

          (line[i - 1] == '-' or Search(r'\boperator\s*$', line[0:i - 1]))):

        continue

      # Pop the stack if there is a matching '<'.  Otherwise, ignore

      # this '>' since it must be an operator.

      if stack:

        if stack[-1] == '<':

          stack.pop()

          if not stack:

            return (i + 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)

  # Did not find end of expression or unbalanced parentheses on this line

  return (-1, stack)

def CloseExpression(clean_lines, linenum, pos):

    return collapsed

def FindEndOfExpressionInLine(line, startpos, stack):

  """Find the position just after the end of current parenthesized expression.

  Args:

    line: a CleansedLines line.

    startpos: start searching at this position.

    stack: nesting stack at startpos.

  Returns:

    On finding matching end: (index just after matching end, None)

    On finding an unclosed expression: (-1, None)

    Otherwise: (-1, new stack at end of this line)

  """

  for i in xrange(startpos, len(line)):

    char = line[i]

    if char in '([{':

      # Found start of parenthesized expression, push to expression stack

      stack.append(char)

    elif char == '<':

      # Found potential start of template argument list

      if i > 0 and line[i - 1] == '<':

        # Left shift operator

        if stack and stack[-1] == '<':

          stack.pop()

          if not stack:

            return (-1, None)

      elif i > 0 and Search(r'\boperator\s*$', line[0:i]):

        # operator<, don't add to stack

        continue

      else:

        # Tentative start of template argument list

        stack.append('<')

    elif char in ')]}':

      # Found end of parenthesized expression.

      #

      # If we are currently expecting a matching '>', the pending '<'

      # must have been an operator.  Remove them from expression stack.

      while stack and stack[-1] == '<':

        stack.pop()

      if not stack:

        return (-1, None)

      if ((stack[-1] == '(' and char == ')') or

          (stack[-1] == '[' and char == ']') or

          (stack[-1] == '{' and char == '}')):

        stack.pop()

        if not stack:

          return (i + 1, None)

      else:

        # Mismatched parentheses

        return (-1, None)

    elif char == '>':

      # Found potential end of template argument list.

      # Ignore "->" and operator functions

      if (i > 0 and

          (line[i - 1] == '-' or Search(r'\boperator\s*$', line[0:i - 1]))):

        continue

      # Pop the stack if there is a matching '<'.  Otherwise, ignore

      # this '>' since it must be an operator.

      if stack:

        if stack[-1] == '<':

          stack.pop()

          if not stack:

            return (i + 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)

  # Did not find end of expression or unbalanced parentheses on this line

  return (-1, stack)

def CloseExpression(clean_lines, linenum, pos):