在c++中,如何知道字符串是否以另一个字符串结束?


当前回答

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);
        }
    );
}

其他回答

std::mismatch方法可以用于从两个字符串的末尾开始向后迭代:

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() );
bool endswith(const std::string &str, const std::string &suffix)
{
    string::size_type totalSize = str.size();
    string::size_type suffixSize = suffix.size();

    if(totalSize < suffixSize) {
        return false;
    }

    return str.compare(totalSize - suffixSize, suffixSize, suffix) == 0;
}

你可以使用string::rfind

基于注释的完整示例:

bool EndsWith(string &str, string& key)
{
size_t keylen = key.length();
size_t strlen = str.length();

if(keylen =< strlen)
    return string::npos != str.rfind(key,strlen - keylen, keylen);
else return false;
}

我的观点是:

bool endsWith(std::string str, std::string suffix)
{
   return str.find(suffix, str.size() - suffix.size()) != string::npos;
}

和上面一样,这是我的解

 template<typename TString>
  inline bool starts_with(const TString& str, const TString& start) {
    if (start.size() > str.size()) return false;
    return str.compare(0, start.size(), start) == 0;
  }
  template<typename TString>
  inline bool ends_with(const TString& str, const TString& end) {
    if (end.size() > str.size()) return false;
    return std::equal(end.rbegin(), end.rend(), str.rbegin());
  }