Question

为什么我的函数不影响我在main()中传递的原始矢量？

void swap2(int &x, int &y)
{
    int temp = x;
    x = y;
    y = temp;

}

void reverse_2(vector<int> &v1)
{
    int j = v1.size() - 1;
    cout << j << endl;
    for (int i = 0; i != v1.size(); ++i)
    {
        swap2(v1[i], v1[j]);
        j--;
    }
}

int main()
{
    vector<int>i1 = { 1,3,5,7,9 }, i2(i1.size());
    i2 = reverse_1(i1);
    for (int i = 0; i != i1.size(); ++i)
    {
        //cout << "First array"<<i1[i] << ","<<endl;
        //cout << "Second array" << i2[i] << "," << endl;
    }
    reverse_2(i1);
    cout << "First array changed" << endl;
    for (int i = 0; i != i1.size(); ++i)
        cout << i1[i] << ",";
    cout << endl;
    return 0;
}

Answer 1

您要将向量中的每个元素反转两次。

第一次将0放在n-1上，然后当您按下i=n-1时，将它们交换回来...

for (int i = 0; i < i1.size() / 2; ++i)

C ++交换功能

1 个答案: