查找字符串是否以C++中的另一个字符串结尾

Question

如何在C++中查找字符串是否以另一个字符串结尾？

Answer 1

只需比较最后n个字符std::string::compare:

#include <iostream>

bool hasEnding (std::string const &fullString, std::string const &ending) {
    if (fullString.length() >= ending.length()) {
        return (0 == fullString.compare (fullString.length() - ending.length(), ending.length(), ending));
    } else {
        return false;
    }
}

int main () {
    std::string test1 = "binary";
    std::string test2 = "unary";
    std::string test3 = "tertiary";
    std::string test4 = "ry";
    std::string ending = "nary";

    std::cout << hasEnding (test1, ending) << std::endl;
    std::cout << hasEnding (test2, ending) << std::endl;
    std::cout << hasEnding (test3, ending) << std::endl;
    std::cout << hasEnding (test4, ending) << std::endl;

    return 0;
}

Answer 2

使用此功能:

inline bool ends_with(std::string const & value, std::string const & ending)
{
    if (ending.size() > value.size()) return false;
    return std::equal(ending.rbegin(), ending.rend(), value.rbegin());
}

Answer 3

使用boost::algorithm::ends_with(参见http://www.boost.org/doc/libs/1_34_0/doc/html/boost/algorithm/ends_with.html):

#include <boost/algorithm/string/predicate.hpp>

// works with const char* 
assert(boost::algorithm::ends_with("mystring", "ing"));

// also works with std::string
std::string haystack("mystring");
std::string needle("ing");
assert(boost::algorithm::ends_with(haystack, needle));

std::string haystack2("ng");
assert(! boost::algorithm::ends_with(haystack2, needle));

Answer 4

注意,从c ++ 20 std :: string开始最终将提供starts_with和ends_with.似乎有可能c ++中的c ++ 30字符串可能最终变得可用,如果你不是从遥远的未来读取它,你可以使用这些startsWith/endsWith:

#include <string>

static bool endsWith(const std::string& str, const std::string& suffix)
{
    return str.size() >= suffix.size() && 0 == str.compare(str.size()-suffix.size(), suffix.size(), suffix);
}

static bool startsWith(const std::string& str, const std::string& prefix)
{
    return str.size() >= prefix.size() && 0 == str.compare(0, prefix.size(), prefix);
}

和一些额外的帮助重载:

static bool endsWith(const std::string& str, const char* suffix, unsigned suffixLen)
{
    return str.size() >= suffixLen && 0 == str.compare(str.size()-suffixLen, suffixLen, suffix, suffixLen);
}

static bool endsWith(const std::string& str, const char* suffix)
{
    return endsWith(str, suffix, std::string::traits_type::length(suffix));
}

static bool startsWith(const std::string& str, const char* prefix, unsigned prefixLen)
{
    return str.size() >= prefixLen && 0 == str.compare(0, prefixLen, prefix, prefixLen);
}

static bool startsWith(const std::string& str, const char* prefix)
{
    return startsWith(str, prefix, std::string::traits_type::length(prefix));
}

IMO,c ++字符串显然功能失调,并没有在现实世界的代码中使用.但是希望这至少会变得更好.

Answer 5

我知道C++的问题,但是如果有人需要一个好的'老式C函数来做到这一点:

/*  returns 1 iff str ends with suffix  */
int str_ends_with(const char * str, const char * suffix) {

  if( str == NULL || suffix == NULL )
    return 0;

  size_t str_len = strlen(str);
  size_t suffix_len = strlen(suffix);

  if(suffix_len > str_len)
    return 0;

  return 0 == strncmp( str + str_len - suffix_len, suffix, suffix_len );
}

Answer 6

std::mismatch当用于从两个字符串的末尾向后迭代时,该方法可以用于此目的:

const string sNoFruit = "ThisOneEndsOnNothingMuchFruitLike";
const string sOrange = "ThisOneEndsOnOrange";

const string sPattern = "Orange";

assert( mismatch( sPattern.rbegin(), sPattern.rend(), sNoFruit.rbegin() )
          .first != sPattern.rend() );

assert( mismatch( sPattern.rbegin(), sPattern.rend(), sOrange.rbegin() )
          .first == sPattern.rend() );

Answer 7

在我看来最简单的,C++解决方案是:

bool endsWith(const string& s, const string& suffix)
{
    return s.rfind(suffix) == (s.size()-suffix.size());
}

Answer 8

我们a是一个字符串,b您要查找的字符串.使用a.substr获得的最后n个字符a,并把它们比作B(其中n是长度b)

或使用std::equal(包括<algorithm>)

例如:

bool EndsWith(const string& a, const string& b) {
    if (b.size() > a.size()) return false;
    return std::equal(a.begin() + a.size() - b.size(), a.end(), b.begin());
}

Answer 9

使用 std::equal 算法进行<algorithms>反向迭代：

\n

std::string LogExt = ".log";\nif (std::equal(LogExt.rbegin(), LogExt.rend(), filename.rbegin())) {\n   \xe2\x80\xa6\n}\n

\n

从C++20开始ends_with引入。

\n

Answer 10

让我用不区分大小写的版本扩展Joseph 的解决方案（在线演示）

#include <string>
#include <cctype>

static bool EndsWithCaseInsensitive(const std::string& value, const std::string& ending) {
    if (ending.size() > value.size()) {
        return false;
    }
    return std::equal(ending.crbegin(), ending.crend(), value.crbegin(),
        [](const unsigned char a, const unsigned char b) {
            return std::tolower(a) == std::tolower(b);
        }
    );
}