用：int* findmin(int *minIndex, int a[], int size=10); double* findmin(int *minIndex, double a[], int size=10);函数原型编写出一个findmin函数，能寻找数组中的最小元素

可以使用函数重载，根据传入的数组类型调用对应的函数，来编写一个能寻找数组中最小元素的`findmin`函数，代码如下：

#include <iostream>
using namespace std;

int* findmin(int *minIndex, int a[], int size=10);
double* findmin(int *minIndex, double a[], int size=10);

int main() {
    int arr1[] = {5, 3, 8, 1, 9, 2};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int minIndex1;
    int* ptr1 = findmin(&minIndex1, arr1, n1);
    cout << "The minimum value in the integer array is: " << *ptr1 << ", and its index is: " << minIndex1 << endl;
    
    double arr2[] = {5.1, 3.2, 8.5, 1.2, 9.7, 2.4};
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    int minIndex2;
    double* ptr2 = findmin(&minIndex2, arr2, n2);
    cout << "The minimum value in the double array is: " << *ptr2 << ", and its index is: " << minIndex2 << endl;
    
    return 0;
}

int* findmin(int *minIndex, int a[], int size) {
    int min_val = a[0];
    *minIndex = 0;
    for (int i = 1; i < size; i++) {
        if (a[i] < min_val) {
            min_val = a[i];
            *minIndex = i;
        }
    }
    return &a[*minIndex];
}

double* findmin(int *minIndex, double a[], int size) {
    double min_val = a[0];
    *minIndex = 0;
    for (int i = 1; i < size; i++) {
        if (a[i] < min_val) {
            min_val = a[i];
            *minIndex = i;
        }
    }
    return &a[*minIndex];
}

运行结果为：

The minimum value in the integer array is: 1, and its index is: 3
The minimum value in the double array is: 1.2, and its index is: 3

注意，这里为了方便输出最小元素的索引，将 `minIndex` 定义为指针类型，传入函数时需要使用 `&` 取地址符号。函数内部通过修改 `minIndex` 指向的地址来返回最小元素的索引。