本文介绍了C ++令牌字符串的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
例如,如果用户输入forty_five,那么我们将使用一个简单的方法来对字符串输入进行标记化,而不使用Boost等非默认库。
解决方案要将一个字符串转换为一个令牌的向量线程安全):
std :: vector< std :: string& inline StringSplit(const std :: string& source,const char * delimiter =,bool keepEmpty = false) { std :: vector< std :: string&结果; size_t prev = 0; size_t next = 0; while((next = source.find_first_of(delimiter,prev))!= std :: string :: npos) { if(keepEmpty || != 0)) { results.push_back(source.substr(prev,next - prev)); } prev = next + 1; } if(prev< source.size()) { results.push_back(source.substr(prev)); } 返回结果; }
I'm looking for a simple way to tokenize string input without using non default libraries such as Boost, etc.
For example, if the user enters forty_five, I would like to seperate forty and five using the _ as the delimiter.
解决方案To convert a string to a vector of tokens (thread safe):
std::vector<std::string> inline StringSplit(const std::string &source, const char *delimiter = " ", bool keepEmpty = false) { std::vector<std::string> results; size_t prev = 0; size_t next = 0; while ((next = source.find_first_of(delimiter, prev)) != std::string::npos) { if (keepEmpty || (next - prev != 0)) { results.push_back(source.substr(prev, next - prev)); } prev = next + 1; } if (prev < source.size()) { results.push_back(source.substr(prev)); } return results; }
更多推荐
C ++令牌字符串
发布评论