中綴到后綴的轉換：堆棧無法識別錯誤

Question

我是使用堆棧的初學者，所以我一直在嘗試對其進行不同的練習。 我正在嘗試轉換中綴-> 后綴。 xcode 調試器說“使用類模板‘堆棧’需要模板參數’。這是我的代碼。

#include<iostream>
#include<stack>
#include<string>

using namespace std;


bool IsOperand(char ch)
{
    if ((ch >= '0' && ch <= '9') || (ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z')) {
        return true;
    }
    return false;
}

bool IsOperator(char C)
{
    if (C == '+' || C == '-' || C == '*' || C == '/' || C == '^') {
        return true;
    }
    return false;
}
bool IsLeftParenthesis(char ch)
{
    if (ch == '(') {
        return true;
    }
    return false;
}

bool IsRightParenthesis(char ch)
{
    if (ch == ')') {
        return true;
    }
    return false;
}

bool Flag(char ch)
{
    if (!IsOperand(ch) || !IsOperator(ch) || !IsLeftParenthesis(ch) || !IsRightParenthesis(ch)) {
        return false;
    }
    return true;
}

int IsRightAssociative(char op)
{
    if (op == '^') {
        return true;
    }
    return false;
}
int GetOperatorWeight(char op)
{
    int weight = -1;
    switch (op) {
        case '+':
        case '-':
            weight = 1;
            break;
        case '*':
        case '/':
            weight = 2;
            break;
        case '^':
            weight = 3;
            break;
    }
    return weight;
}

bool HasHigherPrecedence(char op1, char op2)
{
    int op1Weight = GetOperatorWeight(op1);
    int op2Weight = GetOperatorWeight(op2);
    // If operators have equal precedence, return true if they are left associative.
    // BUT REMEMBER...return false, if right associative.
    // if operator is left-associative, left one should be given priority.
    if (op1Weight == op2Weight) {
        if (IsRightAssociative(op1)) {
            return false;
        }
        else {
            return true;
        }
    }
    return op1Weight > op2Weight ? true : false;
}

string InfixToPostfix(string expression)
{

    stack S;
    string postfix = "";
    for (auto& elem : expression) {
        if (Flag(elem)) {
            continue;
        }
        // If character is operator, pop two elements from stack, perform operation and push the result back.
        else if (IsOperator(elem)) {
            while (!S.empty() && S.top() != '(' && HasHigherPrecedence(S.top(), elem)) {
                postfix += S.top();
                S.pop();
            }
            S.push(elem);
        }
        else if (IsOperand(elem)) {
            postfix += elem;
        }
        else if (elem == '(') {
            S.push(elem);
        }
        else if (elem == ')') {
            while (!S.empty() && S.top() != '(') {
                postfix += S.top();
                S.pop();
            }
            S.pop();
        }
    }

    while (!S.empty()) {
        postfix += S.top();
        S.pop();
    }

    return postfix;
}

int main()
{
    // std::string expression = "54/(5^2)+(6^2^3)";
    std::string expression = "A+(BC-(D/E^F)G)H";
    std::string postfix = InfixToPostfix(expression);
    std::cout << "Output = " << postfix << "\n";
}

這里特別是發生錯誤的地方。

string InfixToPostfix(string expression)
{

    stack S;

它說

Stack S -> 使用類模板'stack'需要模板參數'

Answer 1

Stack 是一個容器，您需要指定容器的類型，例如：

stack <int> S;

或者在你的情況下它是一堆char ：

stack <char> S;