c++ - 修改调车场算法 (c++)

Question

我有一个正常工作的调车场算法，但我注意到一个特殊的怪癖:

1 + ( 3 * ( 4 + 5 ) )

parses correctly to

1 3 4 5 + * +,

but

1 + (3 * (4 + 5))

fails, and parses to

1 * + 5)) +

I want to get it to parse the second problem properly, so that the result is the same as the first. How can I accomplish this?

Notes: I derived my algorithm from wikipedia: http://en.wikipedia.org/wiki/Shunting-yard_algorithm#The_algorithm_in_detail

My algorithm code is:

string switchingYard(string input)
{
stringstream io(input);
ProcessStack switch_stack;
vector<string> out;
while(io.good())
{
    string token;
    io >> token;
    if(isdigit(token[0]) || (token[0] == '.' && isdigit(token[1]))
        || ( (token[0] == '-' && isdigit(token[1])) || (token[0] == '-' && token[1] == '.' && isdigit(token[2])) ) )
    {
        out.push_back(token);
    }


    if(isFunctionToken(token))
    {
        switch_stack.pushNode(token);
    }

    if(isArgSeparator(token[0]))
    {
        bool mismatch_parens = false;
        do{
            if(switch_stack.length() == 1)
            {
                if(switch_stack.peekChar() != '(')
                {
                    mismatch_parens = true;
                    break;
                }
            }
            string opPop = switch_stack.popNode();
            out.push_back(opPop);
        }while(switch_stack.peekChar() != '(');
        if(mismatch_parens)
            return "MISMATCH_ERROR";
    }

    if(isOperator(token[0]))
    {
        while(  isOperator(switch_stack.peekChar()) &&
                ((left_assoc(token[0]) && (op_preced(token[0]) == op_preced(switch_stack.peekChar()) )) || (op_preced(token[0]) < op_preced(switch_stack.peekChar())) ) )
        {
            string popped = switch_stack.popNode();
            out.push_back(popped);
        }
        switch_stack.pushNode(token);
    }

    if(token == "(")
        switch_stack.pushNode(token);

    if(token == ")")
    {
        bool mismatch_parens = false;
        while(switch_stack.peekChar() != '(')
        {
            if(switch_stack.length() == 0 || (switch_stack.length() == 1 && switch_stack.peekChar() != '('))
            {
                mismatch_parens = true;
                break;
            }
            string opPop = switch_stack.popNode();
            out.push_back(opPop);
        }
        if(mismatch_parens)
            return "MISMATCH_ERROR";
        string parensPop = switch_stack.popNode();
        if(isFunctionToken(switch_stack.peek()))
        {
            string funcPop = switch_stack.popNode();
            out.push_back(funcPop);
        }

    }
}
while(switch_stack.length() > 0)
{
    if(switch_stack.peekChar() == '(' || switch_stack.peekChar() == ')')
        return "MISMATCH_ERROR";
    string opPop = switch_stack.popNode();
    out.push_back(opPop);
}
string ret;
for(int i = 0; (unsigned)i < out.size(); i++)
{
    ret += out[i];
    if((unsigned)i < out.size()-1)
        ret += " ";
}
cout << "returning:\n" << ret << endl;
return ret;
}

编辑:好的，所以我有了一个想法。因此，当解析器遇到 '(3' 标记时，它会将这两个字符视为一个字符，并丢弃整个字符，但是如果我递归调用函数，传入输入字符串以“3”字符开始？然后我只需要将分流的字符串添加到输出 vector ，并在字符串流上调用忽略! 我说的是进行这些更改:

string switchingYard(string input)

成为

string switchingYard(string input, int depth)

和

if((token[0] == '(' || token[0] == ')') && (isdigit(token[1]) || token[1] == '.' || isOperator(token[1]))
            {
                string shunted_recur = out.push_back(switchingYard(input.substr(io.tellg()+1),depth+1));
            }

被添加到 while 循环的末尾。想法？

Answer 1

你的问题是解析器读取字符串:

io >> token;

在我看来，简单的解决方案是一次只读取一个字符。例如。

char ch;

io >> ch;

我实际上会写一个函数来读取一个标记——它会知道诸如“数字序列是一个数字”之类的东西，并将运算符、括号等分开。它会返回一个对象(类或结构类型)，它包含“元素类型”和“值(如果相关)”- 因此它可以是类型“数字”和值“4711”，或者类型“运算符”，值“+”。

你的分词器需要一个“状态”，其中包括一个“先行”字符(一个字符应该足够了)，这样你就可以在超过数字末尾时停止，然后继续下一次“不再是数字”的角色。