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;
}

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;
}

从c++ 23开始，你可以使用std::string::contains

#include <string>

const auto haystack = std::string("haystack with needles");
const auto needle = std::string("needle");

if (haystack.contains(needle))
{
    // found!
}

还可以使用System命名空间。 然后可以使用contains方法。

#include <iostream>
using namespace System;

int main(){
    String ^ wholeString = "My name is Malindu";

    if(wholeString->ToLower()->Contains("malindu")){
        std::cout<<"Found";
    }
    else{
        std::cout<<"Not Found";
    }
}

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。