指针求n个数中绝对值的最小值

请参考下面代码，a[i]也可以写成*(a + i)，其中，i=0..n-1

int absmin(int *a, int n)
{
    if (!a || n <= 0) return INT_MIN;

    int minNum = abs(a[0]);
    for (int i = 1; i < n; i++)
    {
        minNum = min(minNum, abs(a[i]));
    }
    return minNum;
}
int main()
{
    int a[] = { -20, -40, 10, 50 };
    int r = absmin(a, sizeof(a) / sizeof(a[0]));
    cout << r << endl;
    return 0;
}