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;
}
最佳答案
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。
<支持> 1. 这不是为了有用,只是为了包含两种捕获和的参数。
关于c++ - 递归函数 - 删除特殊字符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33602287/