如何用一个分隔符将一个string语句分开来
例如string a = "aaa bbb ccc"
有没有哪个函数能将他分为aaa bbb ccc3段放到3个string 中的？
[解决办法]
原型
char *strtok(char s[], const char *delim);
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功能
分解字符串为一组字符串。s为要分解的字符串，delim为分隔符字符串。
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说明
strtok()用来将字符串分割成一个个片段。参数s指向欲分割的字符串，参数delim则为分割字符串中包含的所有字符。当strtok()在参数s的字符串中发现参数delim中包涵的分割字符时,则会将该字符改为\0 字符。在第一次调用时，strtok()必需给予参数s字符串，往后的调用则将参数s设置成NULL。每次调用成功则返回指向被分割出片段的指针。
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返回值
从s开头开始的一个个被分割的串。当没有被分割的串时则返回NULL。
所有delim中包含的字符都会被滤掉，并将被滤掉的地方设为一处分割的节点。
[解决办法]
自己实现个split放vector里面吧,毕竟大多数情况下,分割之前你不知道会分成多少个
#include <iostream>
#include <string>
#include <vector>
using namespace std;
void split(vector<string> &v, string s, string delimitor=" ")
{
size_t pos = s.find(delimitor);
size_t last_pos = 0;
v.clear();
while (pos!=string::npos)
{
v.push_back(s.substr(last_pos,pos-last_pos));
last_pos = pos+delimitor.size();
pos = s.find(delimitor,last_pos);
}
if (last_pos<s.size()) v.push_back(s.substr(last_pos));
}
int main()
{
string a = "aaa bbb cde";
vector<string> v;
split(v,a," ");
for (size_t i=0;i<v.size();i++) cout << i << "-" << v[i] << " ";
cout << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
char string[80];
char seps1[3];
char seps2[3];
char *token;
char *zzstrtok (
char *string,
const char *control1,//连续出现时视为中间夹空token
const char *control2 //连续出现时视为中间无空token
)
{
unsigned char *str;
const unsigned char *ctrl1 = (const unsigned char *)control1;
const unsigned char *ctrl2 = (const unsigned char *)control2;
unsigned char map1[32],map2[32];
static char *nextoken;
static char flag=0;
unsigned char c;
int L;
memset(map1,0,32);
memset(map2,0,32);
do {
map1[*ctrl1 >> 3]
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= (1 << (*ctrl1 & 7));
} while (*ctrl1++);
do {
map2[*ctrl2 >> 3]
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= (1 << (*ctrl2 & 7));
} while (*ctrl2++);
if (string) {
if (control2[0]) {
L=strlen(string);
while (1) {
c=string[L-1];
if (map2[c >> 3] & (1 << (c & 7))) {
L--;
string[L]=0;
} else break;
}
}
if (control1[0]) {
L=strlen(string);
c=string[L-1];
if (map1[c >> 3] & (1 << (c & 7))) {
string[L]=control1[0];
string[L+1]=0;
}
}
str=(unsigned char *)string;
}
else str=(unsigned char *)nextoken;
string=(char *)str;
while (1) {
if (0==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
string++;
str++;
} else {
flag=1;
str++;
}
} else if (1==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=0;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=2;
break;
} else str++;
} else {//2==flag
if (!*str) return NULL;
if (map1[*str >> 3] & (1 << (*str & 7))) {
str++;
string=(char *)str;
flag=0;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
str++;
string=(char *)str;
} else {
string=(char *)str;
str++;
flag=1;
}
}
}
nextoken=(char *)str;
if (string==(char *)str) return NULL;
else return string;
}
void main()
{
strcpy(string,"A \tstring\t\tof ,,tokens\n\nand some more tokens, ");
strcpy(seps1,",\n");strcpy(seps2," \t");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
China
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"
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");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"");strcpy(seps2,"
[解决办法]
");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"
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");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,a,,b,,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,,");
strcpy(seps1,",");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
}
//
//[A string of ,,tokens
//
//and some more tokens,]
//Tokens:
// <A>, <string>, <of>, <>, <tokens>, <>, <and>, <some>, <more>, <tokens>, <>,
//[1234
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LIYI
[解决办法]
China
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <China>, <010>, <201110260000>, <OK>,
//[1234
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LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <010>, <201110260000>, <OK>,
//[1234
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LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <>, <010>, <201110260000>, <OK>,
//[a]
//Tokens:
// <a>,
//[a,b]
//Tokens:
// <a>, <b>,
//[a,,b]
//Tokens:
// <a>, <>, <b>,
//[,a]
//Tokens:
// <>, <a>,
//[a,]
//Tokens:
// <a>, <>,
//[,a,,b]
//Tokens:
// <>, <a>, <>, <b>,
//[,,a,,b,,]
//Tokens:
// <>, <>, <a>, <>, <b>, <>, <>,
//[,]
//Tokens:
// <>, <>,
//[,,]
//Tokens:
// <>, <>, <>,
//[,,,]
//Tokens:
// <>, <>, <>, <>,
