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场景:
1.在存储数据时有时接口需要合并字符串值,并以某些特殊字符来合并部分,到需要的时候再分割它。如一些数值,人名等。
2.C++有strtok,stringstream和find函数来实现分割。可以根据情况调用。
#include <stdlib.h> #include <string.h> #include #include <iostream> #include <sstream> #include <vector> using namespace std; void TestStrtok() { //1.非线程安全的,如果多个线程同时调用,会覆盖掉原来的值. //2.支持以字符串分割. //3.源字符串必须是可修改的. char c_str[]="google||twitter||facebook||microsoft||apple||ibm||"; const char* delim = "||"; char* result = strtok(c_str,delim); while(result != NULL) { cout << result << endl; result = strtok(NULL,delim); } } void TestGetLineWithStringStream() { //1.线程安全的,但是只能以字符作为分隔符 stringstream ss("google|twitter|facebook|microsoft|apple|ibm|"); string str; while(getline(ss,str,‘|‘)) { cout << str << endl; } } void TestStringFind() { //1.自己实现,线程安全,支持字符串作为分隔符.缺点可能就是代码量多. string str = "google||twitter||facebook||microsoft||apple||ibm||"; const char* delim = "||"; const int len = strlen(delim); size_t index = 0; size_t pos = str.find(delim,index); while(pos != string::npos) { string ss = str.substr(index,pos-index); cout << ss << endl; index = pos+len; pos = str.find(delim,index); } //cout << "is last?" << " index:" << index << " str.length():" << str.length() << endl; if((index+1) < str.length()) { string ss = str.substr(index,str.length() - index); cout << ss << endl; } } int main(int argc, char const *argv[]) { cout << "TestStrtok: " << endl; TestStrtok(); cout << "TestGetLineWithStringStream: " << endl; TestGetLineWithStringStream(); cout << "TestStringFind: " << endl; TestStringFind(); return 0; }
输出:
TestStrtok:
google
twitter
facebook
microsoft
apple
ibm
TestGetLineWithStringStream:
google
twitter
facebook
microsoft
apple
ibm
TestStringFind:
google
twitter
facebook
microsoft
apple
ibm
[Finished in
0
.2s]
char* a[3];
char* buf ="这是第一行\n这是第二行\n这是第三行\n";
我想要用‘\n‘符将buf分割成三段并分别存入a[1],a[2],a[3]中,
请问该怎么做~
#include <stdio.h> #include <string.h> #include <malloc.h> int main() { char *a[3]; char *buf ="这是第一行\n这是第二行\n这是第三行\n"; char *t, *pre = buf; int i = 0, l; while (t = strchr(pre, ‘\n‘)) { if (i >= 3) break; l = t - pre; a[i] = (char *)malloc(l + 1); strncpy(a[i], pre, l); a[i][l] = ‘\0‘; ++i; pre = t + 1; } for (i = 0; i < 3; ++i) { printf("%s\n", a[i]); free(a[i]); } return 0; }
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原文地址:http://www.cnblogs.com/SparkOng/p/5034266.html