c++ - 错误 : No matching overloaded function found

Question

int main()
{
    string str;
    cout << "Enter Infix Expression \n";
    cin >> str;
    cout << "infix:" << str << "\n";
    string postfix = InfixToPostfix(str); // **error cause here**
    cout << "postfix:  " << postfix << "\n\n";

    system("pause");
    return 0;
}

// Function to evaluate Postfix expression and return output
template <class T>
string InfixToPostfix(string& str)
{
    Stack<char> *charStackPtr;
    charStackPtr = new Stack<char>();

    string postfix = ""; // Initialize postfix as empty string.
    for (int i = 0; i< str.length(); i++) {
        // If character is operator, pop two elements from stack, perform operation and push the result back. 
        if (IsOperator(str[i]))
        {
            while (!charStackPtr.empty() && charStackPtr.top() != '(' && HasHigherPrecedence(charStackPtr.top(), str[i]))
            {
                postfix += charStackPtr.top();
                charStackPtr.pop();
            }
            charStackPtr.push(str[i]);
        }
        // Else if character is an operand
        else if (IsOperand(str[i]))
        {
            postfix += str[i];
        }

        else if (str[i] == '(')
        {
            charStackPtr.push(str[i]);
        }

        else if (str[i] == ')')
        {
            while (!charStackPtr.empty() && charStackPtr.top() != '(') {
                postfix += charStackPtr.top();
                charStackPtr.pop();
            }
            charStackPtr.pop();
        }
    }while (!charStackPtr.empty()) {
        postfix += charStackPtr.top();
        charStackPtr.pop();
    }

    delete charStackPtr;
    return postfix;
}

谁能帮我解决为什么我不能运行这个程序，我总是犯 3 个错误:

Error C2672 'InfixToPostfix': no matching overloaded function found

Error C2783 'std::string InfixToPostfix(std::string)': could not deduce template argument for 'T'

E0304 no instance of overloaded function "InfixToPostfix" matches the argument list

Answer 1

template <class T>
string InfixToPostfix(string& str)

这表示该函数接受任何类型 T 作为其参数。如果函数的参数之一是 T 类型的变量，则编译器将能够找到并推断出特定的重载。

i am trying to use the stack template that I created, not from the library

您的堆栈声明为:

Stack<char> *charStackPtr

由于堆栈总是将是 char 类型，因此您不需要为它模板参数 T。解决方案是将其删除。 在变量具有已知类型的函数中使用模板变量不需要函数本身是模板。