下面的方法是可以的, 但是影響性能,有沒有現(xiàn)成的函數(shù)可用,很多函數(shù)都試過了,還是不行l(wèi)ist<CString>::iterator ii;int szie = g_Queue.size();int k =0;CString str;for (ii = g_Queue.begin(); ii != g_Queue.end(); ++ii){k++;if(k==size){str= (CString)*ii;}}
2 回答

HUH函數(shù)
TA貢獻1836條經(jīng)驗 獲得超4個贊
Example // list::rbegin/rend
#include <iostream>
#include <list>
using namespace std;
int main ()
{
list<int> mylist;
for (int i=1; i<=5; i++) mylist.push_back(i);
cout << "mylist contains:";
list<int>::reverse_iterator rit;
for ( rit=mylist.rbegin() ; rit != mylist.rend(); ++rit )
cout << " " << *rit;
cout << endl;
return 0;
}
Output:
mylist contains: 5 4 3 2 1
你直接用
list<CString>::reverse_iterator rit=g_Queue.rbegin() ;
str= (CString)*rit;
就可以了
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