milvus-io / milvus

  return (line, clean_lines.NumLines(), -1)

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 ReverseCloseExpression(clean_lines, linenum, pos):

  return (line, clean_lines.NumLines(), -1)

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 ReverseCloseExpression(clean_lines, linenum, pos):