Java有一个方便的分割方法:
String str = "The quick brown fox";
String[] results = str.split(" ");
在c++中有简单的方法来做到这一点吗?
Java有一个方便的分割方法:
String str = "The quick brown fox";
String[] results = str.split(" ");
在c++中有简单的方法来做到这一点吗?
当前回答
If the maximum length of the input string to be tokenized is known, one can exploit this and implement a very fast version. I am sketching the basic idea below, which was inspired by both strtok() and the "suffix array"-data structure described Jon Bentley's "Programming Perls" 2nd edition, chapter 15. The C++ class in this case only gives some organization and convenience of use. The implementation shown can be easily extended for removing leading and trailing whitespace characters in the tokens.
基本上,可以将分隔符替换为以字符串结束的'\0'字符,并设置指向修改后字符串中的标记的指针。在极端情况下,当字符串仅由分隔符组成时,将得到字符串长度加1个空标记。复制要修改的字符串是可行的。
头文件:
class TextLineSplitter
{
public:
TextLineSplitter( const size_t max_line_len );
~TextLineSplitter();
void SplitLine( const char *line,
const char sep_char = ',',
);
inline size_t NumTokens( void ) const
{
return mNumTokens;
}
const char * GetToken( const size_t token_idx ) const
{
assert( token_idx < mNumTokens );
return mTokens[ token_idx ];
}
private:
const size_t mStorageSize;
char *mBuff;
char **mTokens;
size_t mNumTokens;
inline void ResetContent( void )
{
memset( mBuff, 0, mStorageSize );
// mark all items as empty:
memset( mTokens, 0, mStorageSize * sizeof( char* ) );
// reset counter for found items:
mNumTokens = 0L;
}
};
Implementattion文件:
TextLineSplitter::TextLineSplitter( const size_t max_line_len ):
mStorageSize ( max_line_len + 1L )
{
// allocate memory
mBuff = new char [ mStorageSize ];
mTokens = new char* [ mStorageSize ];
ResetContent();
}
TextLineSplitter::~TextLineSplitter()
{
delete [] mBuff;
delete [] mTokens;
}
void TextLineSplitter::SplitLine( const char *line,
const char sep_char /* = ',' */,
)
{
assert( sep_char != '\0' );
ResetContent();
strncpy( mBuff, line, mMaxLineLen );
size_t idx = 0L; // running index for characters
do
{
assert( idx < mStorageSize );
const char chr = line[ idx ]; // retrieve current character
if( mTokens[ mNumTokens ] == NULL )
{
mTokens[ mNumTokens ] = &mBuff[ idx ];
} // if
if( chr == sep_char || chr == '\0' )
{ // item or line finished
// overwrite separator with a 0-terminating character:
mBuff[ idx ] = '\0';
// count-up items:
mNumTokens ++;
} // if
} while( line[ idx++ ] );
}
使用的场景是:
// create an instance capable of splitting strings up to 1000 chars long:
TextLineSplitter spl( 1000 );
spl.SplitLine( "Item1,,Item2,Item3" );
for( size_t i = 0; i < spl.NumTokens(); i++ )
{
printf( "%s\n", spl.GetToken( i ) );
}
输出:
Item1
Item2
Item3
其他回答
如果你愿意使用C语言,你可以使用strtok函数。在使用它时,您应该注意多线程问题。
我以前只用标准库做了一个lexer/tokenizer。代码如下:
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
using namespace std;
string seps(string& s) {
if (!s.size()) return "";
stringstream ss;
ss << s[0];
for (int i = 1; i < s.size(); i++) {
ss << '|' << s[i];
}
return ss.str();
}
void Tokenize(string& str, vector<string>& tokens, const string& delimiters = " ")
{
seps(str);
// Skip delimiters at beginning.
string::size_type lastPos = str.find_first_not_of(delimiters, 0);
// Find first "non-delimiter".
string::size_type pos = str.find_first_of(delimiters, lastPos);
while (string::npos != pos || string::npos != lastPos)
{
// Found a token, add it to the vector.
tokens.push_back(str.substr(lastPos, pos - lastPos));
// Skip delimiters. Note the "not_of"
lastPos = str.find_first_not_of(delimiters, pos);
// Find next "non-delimiter"
pos = str.find_first_of(delimiters, lastPos);
}
}
int main(int argc, char *argv[])
{
vector<string> t;
string s = "Tokens for everyone!";
Tokenize(s, t, "|");
for (auto c : t)
cout << c << endl;
system("pause");
return 0;
}
我知道这个问题已经有了答案,但我想有所贡献。也许我的解决方案有点简单,但这就是我想到的:
vector<string> get_words(string const& text, string const& separator)
{
vector<string> result;
string tmp = text;
size_t first_pos = 0;
size_t second_pos = tmp.find(separator);
while (second_pos != string::npos)
{
if (first_pos != second_pos)
{
string word = tmp.substr(first_pos, second_pos - first_pos);
result.push_back(word);
}
tmp = tmp.substr(second_pos + separator.length());
second_pos = tmp.find(separator);
}
result.push_back(tmp);
return result;
}
如果在我的代码中有更好的方法,或者有什么错误,请评论。
更新:添加通用分隔符
我只是看了所有的答案,找不到下一个前提条件的解决方案:
没有动态内存分配 不使用boost 不使用正则表达式 c++17标准
这就是我的解
#include <iomanip>
#include <iostream>
#include <iterator>
#include <string_view>
#include <utility>
struct split_by_spaces
{
std::string_view text;
static constexpr char delim = ' ';
struct iterator
{
const std::string_view& text;
std::size_t cur_pos;
std::size_t end_pos;
std::string_view operator*() const
{
return { &text[cur_pos], end_pos - cur_pos };
}
bool operator==(const iterator& other) const
{
return cur_pos == other.cur_pos && end_pos == other.end_pos;
}
bool operator!=(const iterator& other) const
{
return !(*this == other);
}
iterator& operator++()
{
cur_pos = text.find_first_not_of(delim, end_pos);
if (cur_pos == std::string_view::npos)
{
cur_pos = text.size();
end_pos = cur_pos;
return *this;
}
end_pos = text.find(delim, cur_pos);
if (cur_pos == std::string_view::npos)
{
end_pos = text.size();
}
return *this;
}
};
[[nodiscard]] iterator begin() const
{
auto start = text.find_first_not_of(delim);
if (start == std::string_view::npos)
{
return iterator{ text, text.size(), text.size() };
}
auto end_word = text.find(delim, start);
if (end_word == std::string_view::npos)
{
end_word = text.size();
}
return iterator{ text, start, end_word };
}
[[nodiscard]] iterator end() const
{
return iterator{ text, text.size(), text.size() };
}
};
int main(int argc, char** argv)
{
using namespace std::literals;
auto str = " there should be no memory allocation during parsing"
" into words this line and you should'n create any"
" contaner for intermediate words "sv;
auto comma = "";
for (std::string_view word : split_by_spaces{ str })
{
std::cout << std::exchange(comma, ",") << std::quoted(word);
}
auto only_spaces = " "sv;
for (std::string_view word : split_by_spaces{ only_spaces })
{
std::cout << "you will not see this line in output" << std::endl;
}
}
您可以简单地使用正则表达式库并使用正则表达式解决该问题。
使用表达式(\w+)和\1中的变量(或$1,取决于正则表达式的库实现)。