C语言字符串组成和分割问题
\u7528C\u8bed\u8a00\u5206\u5272\u5b57\u7b26\u4e32char szPortsMessage[]="0,1,4,6,7;-1062708991|0,3,4,5,7;-1062708992|";
char delims_switchs[] = "|";
char *result_switchs = NULL;
result_switchs = strtok(szPortsMessage,delims_switchs);
while(result_switchs != NULL) {
printf( "result_switchs is \"%s\"\n",result_switchs);
char* tmp = result_switchs; ---------\u52a0\u4e00\u884c\uff0ctmp\uff1d\u5206\u5272\u540e\u7684\u7b2c\u4e00\u4e2a\u5b57\u7b26\u4e32
result_switchs = strtok(NULL,delims_switchs); result_switchs \u5df2\u7ecf\u7b49\u4e8e\u5206\u5272\u540e\u7684\u7b2c\u4e8c\u4e2a\u5b57\u7b26\u4e32\u4e86
char swinfo[100];
strcpy(swinfo,tmp); \uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\u6362\u6210tmp
char delims_switch[] = ";";
char *result_switch = NULL;
result_switch = strtok(swinfo,delims_switch);
while(result_switch != NULL) {
printf( "result_switch is \"%s\"\n",result_switch);
result_switch = strtok(NULL,delims_switch);
}
}
\u8981\u8fbe\u5230\u8fd9\u6837\u7684\u6548\u679c\uff0c\u7528sscanf\u770b\u6765\u662f\u641e\u4e0d\u5b9a\u4e86\uff0c\u53ea\u80fd\u81ea\u5df1\u5199\u51fd\u6570\u3002
\u4e0b\u9762\u7684\u51fd\u6570\u770b\u662f\u5426\u6ee1\u8db3\u9700\u6c42\uff0c\u4f60\u53ef\u4ee5\u5728\u5168\u5c40\u53d8\u91cfstr[]\u91cc\u52a0\u4efb\u610f\u591a\u7684\u8def\u5f84\uff0c\u4f46\u6709\u4e00\u70b9\u6ce8\u610f\u7684\u662f\uff0c\u5b57\u7b26\\u5728\u4e32\u91cc\u5fc5\u987b\u8f6c\u4e49\u4e3a\\\uff0c\u8def\u5f84\u957f\u5ea6\u4e0d\u8d85\u8fc7128\u4f4d\u5373\u53ef\u3002
#include
#include
#include
char str[] = "clamscan c:\\Documents and Settings d:\\test e:\\a and b f:\\c and d";
void main()
{
int i = 0, j = 0;
char (*pStr)[128] = NULL;
char *s = str;
char *ss = NULL;
while (1)
{
pStr = (char (*)[128])realloc(pStr, (i+1)*(128));
memset(*(pStr+i), 0, 128);
ss = strchr(s, ':');
if (i == 0)
{
if (ss == NULL)
{
strcpy(*(pStr+i), s);
i++;
break;
}
memcpy(*(pStr+i), s, ss-s-2);
s = ss+1;
i++;
continue;
}
if (ss == NULL)
{
strcpy(*(pStr+i), s-2);
i++;
break;
}
memcpy(*(pStr+i), s-2, ss-s);
s = ss+1;
i++;
}
printf("After despare, the strings are:\n");
for (j = 0; j < i; j++)
{
printf("[%s]\n", *(pStr+j));
}
free(pStr);
}
\u8f93\u51fa\u6548\u679c\uff1a
char buf[100];
sprintf(buf, "%d,%d,%d,%d,%d", a, b, c, d, e);
这样buf数组中就存得是"10,20,30,40,50"这个字符串了。
想分开的话，定义一个指针p，让这个指针从buf[0] 开始指，让p ++，每次直到逗号的时候就跳出，如果不是逗号的话，就将当前指针指向的这个值放到另一个字符数组buf1中，最后将buf1转成数字，用atoi函数，这时候会得到这个数1020304050，用这个数对100取余。得到50，然后除以100再对100取余，得到40，同理，就会得到相应的数字了，把这些数字存入相应的变量就行。
不明白继续追问
#include <stdio.h>
int main()
{
int a,b,c,d,e;
int f,g,h,i,k;
char buf[20];
a = 10;
b = 20;
c = 30;
d = 40;
e = 50;
sprintf(buf,"%02d,%02d,%02d,%02d,%02d",a,b,c,d,e);
sscanf(buf,"%d,%d,%d,%d,%d",&f,&g,&h,&i,&k);
getchar();
return 0;
}
#include<sstream>
#include<iostream>
#include<string>
using naemspace std;
int main()
{
int i;
int a[5];
string temp,str;
stringstream ss;
for(i=0;i<5;i++)
{
ss<<a[i];
ss>>temp;
if(i<4)
temp+=',';
str+=temp;
ss.clear();
}//////////////////str=10,20,30,40,50
int b[5];
for(i=0;i<5;i++)
{
ss<<str[2*i];
ss>>b[i];
ss.clear();
}
return 0;
}
完成！！！
其实这道题最好用sscanf，你可以百度一下~
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