這是從網(wǎng)上看到的源代碼：（感覺太麻煩，請(qǐng)問哪位大哥能指教我一個(gè)簡(jiǎn)單一點(diǎn)全面一點(diǎn)的分割方法嗎？十分期待）
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
typedef basic_string<char>::size_type S_T;
static const S_T npos = -1;
/**/////trim指示是否保留空串，默認(rèn)為保留。tok可以為任意多個(gè)字符vector<string> tokenize(const string& src, string tok,
bool trim=false, string null_subst="")
{
if( src.empty() || tok.empty() )
throw "tokenize: empty string\0";
vector<string> v;
S_T pre_index = 0, index = 0, len = 0;
while( (index = src.find_first_of(tok, pre_index)) !=npos )
{
if( (len = index-pre_index)!=0 )
v.push_back(src.substr(pre_index, len));
else if(trim==false)
v.push_back(null_subst);
pre_index = index+1;
}
string endstr = src.substr(pre_index);
if( trim==false )
v.push_back( endstr.empty()?null_subst:endstr );
else if( !endstr.empty() )
v.push_back(endstr);
return v;
}
//delimit為一個(gè)字符，嚴(yán)格分割
vector<string> split(const string& src, string delimit, string null_subst="")
{
if( src.empty() || delimit.empty() )
throw "split:empty string\0";
vector<string> v;
S_T deli_len = delimit.size();
long index = npos, last_search_position = 0;
while( (index=src.find(delimit,
last_search_position))!=npos )
{
if(index==last_search_position)
v.push_back(null_subst);
else
v.push_back( src.substr(last_search_position, index-
last_search_position) );
last_search_position = index + deli_len;
}
string last_one = src.substr(last_search_position);
v.push_back( last_one.empty()? null_subst:last_one );
return v;
}
//測(cè)試如下：
int main(int argc, char* argv[])
{
string src = ",ab,cde;,,fg,," ;
string tok = ",;" ;
vector<string> v1 = tokenize(src, tok ,true);
vector<string> v2 = tokenize(src, tok ,false,
"<null>");
cout<<"-------------v1:"<<endl;
for(int i=0; i<v1.size();i++)
{
cout<<v1[i]<<endl;
}
cout<<"-------------v2:"<<endl;
for(int j=0; j<v2.size();j++)
{
cout<<v2[j].c_str()<<endl;
}
try{
string s = "1;2;3;4";
string del = ";";//"###";
vector<string> v3 = split(s, del, "<null>");
cout<<"-------------v3:"<<endl;
for(int k=0; k<v3.size();k++)
{
cout<<v3[k].c_str()<<endl;
}
}
catch (char *s) {
cout<<s<<endl;
}
return 0;
}
posted on 2009-08-07 14:24
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