Function FindEndOfExpressionInLine

cpplint.py:2195–2270

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

(line, startpos, stack)

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 range(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 re.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 re.search(r"\boperator\s*$", line[0 : i - 1])):
                continue