注意:我知道这个问题需要一个函数，这意味着用户试图找到一些更简单的东西。但我还是把它贴出来，以防有人觉得有用。

使用后缀自动机的方法。它接受一个字符串(干草堆)，然后你可以输入成千上万的查询(针)，并且响应将非常快，即使干草堆和/或针是非常长的字符串。

阅读此处使用的数据结构:https://en.wikipedia.org/wiki/Suffix_automaton

#include <bits/stdc++.h>

using namespace std;

struct State {
  int len, link;
  map<char, int> next;
};

struct SuffixAutomaton {
  vector<State> st;
  int sz = 1, last = 0;

  SuffixAutomaton(string& s) {
    st.assign(s.size() * 2, State());
    st[0].len = 0;
    st[0].link = -1;
    for (char c : s) extend(c);
  }

  void extend(char c) {
    int cur = sz++, p = last;
    st[cur].len = st[last].len + 1;
    while (p != -1 && !st[p].next.count(c)) st[p].next[c] = cur, p = st[p].link;
    if (p == -1)
      st[cur].link = 0;
    else {
      int q = st[p].next[c];
      if (st[p].len + 1 == st[q].len)
        st[cur].link = q;
      else {
        int clone = sz++;
        st[clone].len = st[p].len + 1;
        st[clone].next = st[q].next;
        st[clone].link = st[q].link;
        while (p != -1 && st[p].next[c] == q) st[p].next[c] = clone, p = st[p].link;

        st[q].link = st[cur].link = clone;
      }
    }
    last = cur;
  }
};

bool is_substring(SuffixAutomaton& sa, string& query) {
  int curr = 0;

  for (char c : query)
    if (sa.st[curr].next.count(c))
      curr = sa.st[curr].next[c];
    else
      return false;

  return true;
}

// How to use:
// Execute the code
// Type the first string so the program reads it. This will be the string
// to search substrings on.
// After that, type a substring. When pressing enter you'll get the message showing the
// result. Continue typing substrings.
int main() {
  string S;
  cin >> S;

  SuffixAutomaton sa(S);

  string query;
  while (cin >> query) {
    cout << "is substring? -> " << is_substring(sa, query) << endl;
  }
}

如果字符串的大小相对较大(数百字节或更多)，并且c++17可用，您可能需要使用Boyer-Moore-Horspool搜索器(示例来自cppreference.com):

#include <iostream>
#include <string>
#include <algorithm>
#include <functional>

int main()
{
    std::string in = "Lorem ipsum dolor sit amet, consectetur adipiscing elit,"
                     " sed do eiusmod tempor incididunt ut labore et dolore magna aliqua";
    std::string needle = "pisci";
    auto it = std::search(in.begin(), in.end(),
                   std::boyer_moore_searcher(
                       needle.begin(), needle.end()));
    if(it != in.end())
        std::cout << "The string " << needle << " found at offset "
                  << it - in.begin() << '\n';
    else
        std::cout << "The string " << needle << " not found\n";
}

如果不想使用标准库函数，下面是一种解决方案。

#include <iostream>
#include <string>

bool CheckSubstring(std::string firstString, std::string secondString){
    if(secondString.size() > firstString.size())
        return false;

    for (int i = 0; i < firstString.size(); i++){
        int j = 0;
        // If the first characters match
        if(firstString[i] == secondString[j]){
            int k = i;
            while (firstString[i] == secondString[j] && j < secondString.size()){
                j++;
                i++;
            }
            if (j == secondString.size())
                return true;
            else // Re-initialize i to its original value
                i = k;
        }
    }
    return false;
}

int main(){
    std::string firstString, secondString;

    std::cout << "Enter first string:";
    std::getline(std::cin, firstString);

    std::cout << "Enter second string:";
    std::getline(std::cin, secondString);

    if(CheckSubstring(firstString, secondString))
        std::cout << "Second string is a substring of the frist string.\n";
    else
        std::cout << "Second string is not a substring of the first string.\n";

    return 0;
}

从这个网站上的这么多答案中，我没有找到一个明确的答案，所以在5-10分钟内我自己找到了答案。 但这可以在两种情况下实现:

要么你知道你在字符串中搜索的子字符串的位置 要么你不知道它的位置，然后逐字符搜索它……

所以，让我们假设我们在字符串“abcde”中搜索子字符串“cd”，我们使用c++中最简单的substr内置函数

1:

#include <iostream>
#include <string>

    using namespace std;
int i;

int main()
{
    string a = "abcde";
    string b = a.substr(2,2);    // 2 will be c. Why? because we start counting from 0 in a string, not from 1.

    cout << "substring of a is: " << b << endl;
    return 0;
}

2:

#include <iostream>
#include <string>

using namespace std;
int i;

int main()
{
    string a = "abcde";

    for (i=0;i<a.length(); i++)
    {
        if (a.substr(i,2) == "cd")
        {
        cout << "substring of a is: " << a.substr(i,2) << endl;    // i will iterate from 0 to 5 and will display the substring only when the condition is fullfilled 
        }
    }
    return 0;
}

使用std::string::find如下所示:

if (s1.find(s2) != std::string::npos) {
    std::cout << "found!" << '\n';
}

注意:如果s2是s1的子字符串，则打印"found!"， s1和s2都是std::string类型。

这是一个简单的函数

bool find(string line, string sWord)
{
    bool flag = false;
    int index = 0, i, helper = 0;
    for (i = 0; i < line.size(); i++)
    {
        if (sWord.at(index) == line.at(i))
        {
            if (flag == false)
            {
                flag = true;
                helper = i;
            }
            index++;
        }
        else
        {
            flag = false;
            index = 0;
        }
        if (index == sWord.size())
        {
            break;
        }
    }
    if ((i+1-helper) == index)
    {
        return true;
    }
    return false;
}