想把一个十进制改成别的进制的数，但就是运行不成功，问题在哪里？？？跪求！各位爸爸！

Question

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
void print(int num, int base);
long powed(int base,int n);
main()
{
    int x = 20;
    int y = 8;
    print(x,y);
}


void print(int num, int base)
{
    int fine = 0;
    int counter = 0;
    cout << "Now decimal system is " << num << endl;
    vector<int>arr;
    int fin = 0;
    while(num!=0)
    {
        fine = num%base;
        num = num/base;
        arr[counter] = fine;
        counter++;
    }
    for(size_t i=counter-1;i>=0;i--)
        fin += powed(base,i)*arr[i];
    cout << fin;
}
long powed(int base,int n)
{
    long sum = 1;
    for(size_t i=0;i<n;i++)
        sum = sum * base;
    return sum;
}

Answer 1

1.vector使用不当，vectorarr;，这一句没有指定arr的大小，所以默认是0，后面用[]访问的时候就越界了，
由于预先不能确定转换后的位数，所以改成pushback（），
while循环执行完后arr数组里就是转换后的数，但是是逆序的，需要从后往前cout，
2.powed函数不知道要干什么，是个死循环，也会导致越界

void print(int num, int base);
int main()
{
    int x = 20;
    int y = 8;
    print(x, y);
    return 0;
}


void print(int num, int base)
{
    int fine = 0;
    int counter = 0;
    cout << "Now decimal system is " << num << endl;
    vector<int>arr;
    int fin = 0;
    while (num != 0)
    {
        fine = num%base;
        num = num / base;
        arr.push_back(fine);
    }
    for (int i = arr.size() - 1; i >= 0; --i)
        cout << arr[i];
}