c++ - 递归函数 - 删除特殊字符

Question

我需要一个函数来删除字符串中的特殊字符。我正在创建一个程序，它接受一个句子并计算元音并确定它是否是回文。但如果有的话，我需要删除特殊字符。我使用的是 lambda，但它与我使用的编译器不兼容，而这正是我的教授希望我们的程序编译的地方。因此，如果有人有我可以使用的其他功能，我将非常感激。这些是我得到的错误: 错误:预期的主表达式在 â[â 标记之前 错误:预期在 â]â 标记之前的主表达式 错误:期望主表达式在 âcharâ 之前我评论了上面错误所在的行。

 #include<iostream> 
    #include <cmath>
    #include <algorithm>


    using namespace std;

    //Create a structure called Sentence
    struct Sentence
    {
        int CountVowels(string , int);

        public:
        Sentence (string);
        bool isPal(string , int);
        void Print();
        string s;
        int numVowel;
        int length;
        //~Sentence();

    };

    Sentence :: Sentence (string b)
    {
        s = b;
        length = 0;
        numVowel = CountVowels(s, 0);
    }

    //Count Vowels function using recursion 
    int Sentence :: CountVowels(string myWord, int startindex)
    {
        length ++;
        int pandi; 

        if(myWord[startindex])
        {
            if (myWord[startindex] != 'a' && myWord[startindex] != 'e' && myWord[startindex] != 'i' && myWord[startindex] != 'o' && myWord[startindex] != 'u')
            {
                pandi = 0;
            }
        else pandi = 1;
        return pandi + CountVowels(myWord, startindex + 1);
        } 
        return 0;
    }

    // Check if it palindorme using recursion 
    bool Sentence :: isPal(string myWord, int size)
    {
        int r = myWord.size() - size;
        int t = size - 1;

        //size = r will be true whenn the size of the string is even and the 2 middle characters have been checked
        if (size == r || r == t)

            return true;
        //r = t will be true when the size of the string is odd and the two characters on either side of the middle character have been checked



        if (tolower(myWord[r]) != tolower(myWord[t]))

            return false;


        return isPal(myWord, -- size);
    }

    //Display the sentence 
    void Sentence :: Print()
    {
        cout << s [-- length];
        if (length == 0)
        {
            cout << "" << endl;
            return;

        }
        Print ();
    }

    //Main function 

    int main ()
    {
        //Holds user sentence 
        string userW;

        //Ask user to enter a sentence 
        cout << "Enter a sentence: \n";
        getline(cin, userW);
        //Removes special characters 
        //This is where the ERRORS are 
        userW.erase(remove_if(userW.begin(), userW.end(), [](char c) 
        {return !isalpha(c); }), userW.end());

        //Creates userSent under Sentence 
        Sentence userSent(userW);

        //Display the number of vowels
        cout << "The number of vowels in the sentence is " << userSent.numVowel << endl;
        cout << "" << endl;

        //Display if the sentence is a palindrome or not 
        cout << "The sentence" << " is" << 
        (userSent.isPal(userSent.s, userSent.s.size()) ? " Palindrome\n" : " not Palindrome\n");
        cout << "" << endl; 
        //Display the sentence backwards 
        cout << "The sentence spelled backwards is: " << endl;
        userSent.Print();

        return 0;
    }

Answer 1

lambda 只是定义类的简写方式。如果您愿意，您始终可以定义一个没有 lambda 的类似类。例如，像这样的 lambda 表达式¹(其中 a 和 b 的类型为 A 和 B分别):

[&a, =b](char c) { return a.x() + b > c; }

...可以像这样定义为显式类:

class foo { 
    A mutable &a;
    B b;
public:
    foo(A &a, B b) : a(a), b(b) {}

    bool operator()(char c) const { return a.x() + b > c; }
};

这显然要冗长得多(这也是添加 lambda 表达式的原因)，但做的事情大致相同。

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<支持>1. 这不是为了有用，只是为了包含两种捕获和的参数。