#include "stdafx.h"
#include "time.h"
#include "vector"
using namespace std;
#import "C:\Program Files\Common Files\System\ado\msado15.dll" no_namespace rename("EOF","adoEOF")
int _tmain(int argc, _TCHAR* argv[])
{
vector<char* > Arry; static char buff[256]={0}; time_t ff;
Arry.push_back("a1");
time(&ff);_snprintf(buff,sizeof(buff),"%ld",ff);Arry.push_back(buff); Sleep(3000);memset(buff, 0,sizeof(buff));//休眠,清空buff Arry.push_back("a2");
time(&ff);_snprintf(buff,sizeof(buff),"%ld",ff); Arry.push_back(buff);
printf(Arry[0]+_bstr_t("\n"));
printf(Arry[1]+_bstr_t("\n"));
printf(Arry[2]+_bstr_t("\n"));
printf(Arry[3]+_bstr_t("\n")); system("pause");
return 0;
}
中间休眠的3秒钟;为什么 Arry[1] 和 Arry[3] 的值还是一样的
你后面修改了buff的内容,也就修改了前面的值
#include "stdafx.h"
#include "time.h"
#include "vector"
using namespace std;
#import "C:\Program Files\Common Files\System\ado\msado15.dll" no_namespace rename("EOF","adoEOF")
int _tmain(int argc, _TCHAR* argv[])
{
vector<char* > Arry; char* buff;
Arry.push_back("a1");
buff=(char *)malloc(128);
_ltoa(time(NULL),buff,10);
Arry.push_back(buff); Sleep(3000);//休眠,清空buff Arry.push_back("a2");
buff=(char *)malloc(128);
_ltoa(time(NULL),buff,10);
Arry.push_back(buff);
printf(Arry[0]+_bstr_t("\n"));
printf(Arry[1]+_bstr_t("\n"));
printf(Arry[2]+_bstr_t("\n"));
printf(Arry[3]+_bstr_t("\n")); system("pause");
return 0;
}这样好像可以~~
你后面修改了buff的内容,也就修改了前面的值
push_back(buff),就是把这个buff的地址存进去,buff里面的值改了,但这个buff的地址没变的.
所以1,3就一样了.而后面一条程序,buff 分配了2次空间,每次的地址都不一样的.
第一次分配了地址1,里面有值,地址1push_back了.
第二次再分配一次空间,地址不一样了.
再把第二次的地址push_back,所以就成功了.第二条程序,如果是正规地话,一定要记住对
Arry[1],[3]这2个地址进行 free ,否则就有内存泄露了.