我已经按照维基百科里提到的，用C++11实现了调车场算法:

此实现不实现复合函数，参数数目可变的函数和一元运算符。

while there are tokens to be read:
    read a token.
    if the token is a number, then:
        push it to the output queue.
    else if the token is a function then:
        push it onto the operator stack 
    else if the token is an operator then:
        while ((there is a operator at the top of the operator stack)
              and ((the operator at the top of the operator stack has greater precedence)
               or (the operator at the top of the operator stack has equal precedence and the token is left associative))
              and (the operator at the top of the operator stack is not a left parenthesis)):
            pop operators from the operator stack onto the output queue.
        push it onto the operator stack.
    else if the token is a left parenthesis (i.e. "("), then:
        push it onto the operator stack.
    else if the token is a right parenthesis (i.e. ")"), then:
        while the operator at the top of the operator stack is not a left parenthesis:
            pop the operator from the operator stack onto the output queue.
        /* If the stack runs out without finding a left parenthesis, then there are mismatched parentheses. */
        if there is a left parenthesis at the top of the operator stack, then:
            pop the operator from the operator stack and discard it
/* After while loop, if operator stack not null, pop everything to output queue */
if there are no more tokens to read then:
    while there are still operator tokens on the stack:
        /* If the operator token on the top of the stack is a parenthesis, then there are mismatched parentheses. */
        pop the operator from the operator stack onto the output queue.
exit.

正如您所看到的，上面提到的，这个算法不处理一元运算符，假设我有一个！，它比所有其他运算符都强，如果有的话，我应该对我的算法做什么修改呢？

使用！运算符的一些合法示例:

!1
! 1
! (1)
!(  1 + 2)

再加上一个小问题，这个算法能处理像1==2这样的错误语法吗（我认为是的）？