我有一个 std::string
类型的变量。我想检查它是否包含某个std::string
。我该怎么做?
如果找到字符串,是否有一个函数返回true,如果没有,则返回false?
std::basic_string::contains
添加到标准库。
按如下方式使用 std::string::find
:
if (s1.find(s2) != std::string::npos) {
std::cout << "found!" << '\n';
}
注:“找到了!”如果 s2
是 s1
的子字符串,则将打印,s1
和 s2
都是 std::string
类型。
您可以尝试使用 find
函数:
string str ("There are two needles in this haystack.");
string str2 ("needle");
if (str.find(str2) != string::npos) {
//.. found.
}
从 C++23 开始,您可以使用 std::string::contains
#include <string>
const auto haystack = std::string("haystack with needles");
const auto needle = std::string("needle");
if (haystack.contains(needle))
{
// found!
}
std::string::contains()
。
find
告诉您它的位置。
实际上,您可以尝试使用 boost 库,我认为 std::string 没有提供足够的方法来执行所有常见的字符串操作。在 boost 中,您可以只使用 boost::algorithm::contains
:
#include <string>
#include <boost/algorithm/string.hpp>
int main() {
std::string s("gengjiawen");
std::string t("geng");
bool b = boost::algorithm::contains(s, t);
std::cout << b << std::endl;
return 0;
}
find
方法可以解决所讨论的任务。无需引入库依赖项。
contains
是 std::contains
还是 boost::contains
,这似乎是一个很大的缺点。我猜 std::contains 目前不存在,但我不确定假设读者已经记住了 std 中的所有内容是否合理。并且 std::contains
很可能存在于某些未来版本的 c++ 中,这会破坏该程序。
你可以试试这个
string s1 = "Hello";
string s2 = "el";
if(strstr(s1.c_str(),s2.c_str()))
{
cout << " S1 Contains S2";
}
如果该功能对您的系统至关重要,那么使用旧的 strstr
方法实际上是有益的。 algorithm
中的 std::search
方法可能是最慢的。我的猜测是创建这些迭代器需要很多时间。
我用来计时整个事情的代码是
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <random>
#include <chrono>
std::string randomString( size_t len );
int main(int argc, char* argv[])
{
using namespace std::chrono;
const size_t haystacksCount = 200000;
std::string haystacks[haystacksCount];
std::string needle = "hello";
bool sink = true;
high_resolution_clock::time_point start, end;
duration<double> timespan;
int sizes[10] = { 10, 20, 40, 80, 160, 320, 640, 1280, 5120, 10240 };
for(int s=0; s<10; ++s)
{
std::cout << std::endl << "Generating " << haystacksCount << " random haystacks of size " << sizes[s] << std::endl;
for(size_t i=0; i<haystacksCount; ++i)
{
haystacks[i] = randomString(sizes[s]);
}
std::cout << "Starting std::string.find approach" << std::endl;
start = high_resolution_clock::now();
for(size_t i=0; i<haystacksCount; ++i)
{
if(haystacks[i].find(needle) != std::string::npos)
{
sink = !sink; // useless action
}
}
end = high_resolution_clock::now();
timespan = duration_cast<duration<double>>(end-start);
std::cout << "Processing of " << haystacksCount << " elements took " << timespan.count() << " seconds." << std::endl;
std::cout << "Starting strstr approach" << std::endl;
start = high_resolution_clock::now();
for(size_t i=0; i<haystacksCount; ++i)
{
if(strstr(haystacks[i].c_str(), needle.c_str()))
{
sink = !sink; // useless action
}
}
end = high_resolution_clock::now();
timespan = duration_cast<duration<double>>(end-start);
std::cout << "Processing of " << haystacksCount << " elements took " << timespan.count() << " seconds." << std::endl;
std::cout << "Starting std::search approach" << std::endl;
start = high_resolution_clock::now();
for(size_t i=0; i<haystacksCount; ++i)
{
if(std::search(haystacks[i].begin(), haystacks[i].end(), needle.begin(), needle.end()) != haystacks[i].end())
{
sink = !sink; // useless action
}
}
end = high_resolution_clock::now();
timespan = duration_cast<duration<double>>(end-start);
std::cout << "Processing of " << haystacksCount << " elements took " << timespan.count() << " seconds." << std::endl;
}
return 0;
}
std::string randomString( size_t len)
{
static const char charset[] = "abcdefghijklmnopqrstuvwxyz";
static const int charsetLen = sizeof(charset) - 1;
static std::default_random_engine rng(std::random_device{}());
static std::uniform_int_distribution<> dist(0, charsetLen);
auto randChar = [charset, &dist, &rng]() -> char
{
return charset[ dist(rng) ];
};
std::string result(len, 0);
std::generate_n(result.begin(), len, randChar);
return result;
}
在这里,我生成随机 haystacks
并在其中搜索 needle
。 haystack 计数已设置,但每个 haystack 中的字符串长度从开始的 10 增加到最后的 10240。大多数时候,程序实际上花费在生成随机字符串上,但这是意料之中的。
输出是:
Generating 200000 random haystacks of size 10
Starting std::string.find approach
Processing of 200000 elements took 0.00358503 seconds.
Starting strstr approach
Processing of 200000 elements took 0.0022727 seconds.
Starting std::search approach
Processing of 200000 elements took 0.0346258 seconds.
Generating 200000 random haystacks of size 20
Starting std::string.find approach
Processing of 200000 elements took 0.00480959 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00236199 seconds.
Starting std::search approach
Processing of 200000 elements took 0.0586416 seconds.
Generating 200000 random haystacks of size 40
Starting std::string.find approach
Processing of 200000 elements took 0.0082571 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00341435 seconds.
Starting std::search approach
Processing of 200000 elements took 0.0952996 seconds.
Generating 200000 random haystacks of size 80
Starting std::string.find approach
Processing of 200000 elements took 0.0148288 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00399263 seconds.
Starting std::search approach
Processing of 200000 elements took 0.175945 seconds.
Generating 200000 random haystacks of size 160
Starting std::string.find approach
Processing of 200000 elements took 0.0293496 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00504251 seconds.
Starting std::search approach
Processing of 200000 elements took 0.343452 seconds.
Generating 200000 random haystacks of size 320
Starting std::string.find approach
Processing of 200000 elements took 0.0522893 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00850485 seconds.
Starting std::search approach
Processing of 200000 elements took 0.64133 seconds.
Generating 200000 random haystacks of size 640
Starting std::string.find approach
Processing of 200000 elements took 0.102082 seconds.
Starting strstr approach
Processing of 200000 elements took 0.00925799 seconds.
Starting std::search approach
Processing of 200000 elements took 1.26321 seconds.
Generating 200000 random haystacks of size 1280
Starting std::string.find approach
Processing of 200000 elements took 0.208057 seconds.
Starting strstr approach
Processing of 200000 elements took 0.0105039 seconds.
Starting std::search approach
Processing of 200000 elements took 2.57404 seconds.
Generating 200000 random haystacks of size 5120
Starting std::string.find approach
Processing of 200000 elements took 0.798496 seconds.
Starting strstr approach
Processing of 200000 elements took 0.0137969 seconds.
Starting std::search approach
Processing of 200000 elements took 10.3573 seconds.
Generating 200000 random haystacks of size 10240
Starting std::string.find approach
Processing of 200000 elements took 1.58171 seconds.
Starting strstr approach
Processing of 200000 elements took 0.0143111 seconds.
Starting std::search approach
Processing of 200000 elements took 20.4163 seconds.
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(来自 cppreference.com 的示例):
#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";
}
bool contains(const std::string& haystack, const std::string& needle)
。如今,他们提供了一组以一些默默无闻的论文作者命名的拼图,使其看起来更像计算机科学......
如果您不想使用标准库函数,以下是一种解决方案。
#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;
}
CheckSubstring(std::string firstString, std::string secondString)
深度复制传递给函数的两个字符串,这很昂贵,特别是对于需要堆分配的较长字符串。此外,假设您调用 CheckSubstring("XYZab", "ab\0\0")
- while
循环最终会将 a
与 a
、b
与 b
进行比较,即第一个字符串末尾的隐式 NUL 与其次,它将读取超出第一个字符串的缓冲区,具有未定义的行为。要修复使用 for (...
i <= firstString.size() - secondString().size(); ...)`。
#include <algorithm> // std::search
#include <string>
using std::search; using std::count; using std::string;
int main() {
string mystring = "The needle in the haystack";
string str = "needle";
string::const_iterator it;
it = search(mystring.begin(), mystring.end(),
str.begin(), str.end()) != mystring.end();
// if string is found... returns iterator to str's first element in mystring
// if string is not found... returns iterator to mystring.end()
if (it != mystring.end())
// string is found
else
// not found
return 0;
}
using
与仅必需的函数一起使用,而不是将整个命名空间转储到全局空间中。至于@SᴀᴍOnᴇᴌᴀ 评论,我猜该用户没有阅读您代码中的评论。
也很适合使用 std::regex_search
。使搜索更通用的垫脚石。下面是一个带有注释的示例。
//THE STRING IN WHICH THE SUBSTRING TO BE FOUND.
std::string testString = "Find Something In This Test String";
//THE SUBSTRING TO BE FOUND.
auto pattern{ "In This Test" };
//std::regex_constants::icase - TO IGNORE CASE.
auto rx = std::regex{ pattern,std::regex_constants::icase };
//SEARCH THE STRING.
bool isStrExists = std::regex_search(testString, rx);
需要包含#include <regex>
出于某种原因,假设观察到输入字符串类似于“在此示例字符串中查找内容”,并且有兴趣搜索“在此测试中”或“在此示例中”,则可以通过简单地调整模式来增强搜索,如图所示以下。
//THE SUBSTRING TO BE FOUND.
auto pattern{ "In This (Test|Example)" };
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 分钟内我自己想出了答案。但这可以在两种情况下完成:
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位置并搜索它,逐个字符......
因此,假设我们在字符串“abcde”中搜索子字符串“cd”,并且我们使用 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;
}
这是一个简单的功能
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;
}
您还可以使用 System 命名空间。然后你可以使用 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";
}
}
注意:我知道这个问题需要一个函数,这意味着用户正在尝试找到更简单的东西。但我仍然发布它以防有人发现它有用。
使用后缀自动机的方法。它接受一个字符串(haystack),然后您可以输入数十万个查询(needles)并且响应将非常快,即使 haystack 和/或 needle 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;
}
}
我们可以改用这种方法。只是我项目中的一个例子。参考代码。一些额外的东西也包括在内。
查看 if 语句!
/*
Every C++ program should have an entry point. Usually, this is the main function.
Every C++ Statement ends with a ';' (semi-colon)
But, pre-processor statements do not have ';'s at end.
Also, every console program can be ended using "cin.get();" statement, so that the console won't exit instantly.
*/
#include <string>
#include <bits/stdc++.h> //Can Use instead of iostream. Also should be included to use the transform function.
using namespace std;
int main(){ //The main function. This runs first in every program.
string input;
while(input!="exit"){
cin>>input;
transform(input.begin(),input.end(),input.begin(),::tolower); //Converts to lowercase.
if(input.find("name") != std::string::npos){ //Gets a boolean value regarding the availability of the said text.
cout<<"My Name is AI \n";
}
if(input.find("age") != std::string::npos){
cout<<"My Age is 2 minutes \n";
}
}
}