C++函数重载！1：调用MIN函数2定义char，int，float，long数组，3,用MIN函数输出数组中最小值！！

Question

你的答案有错误噢。。。。

Answer 1

在上面那位的基础改了一下，数组不能动态分配，可以直接用宏定义其实，在WinXP SP3 + VS 2005 express下通过

#include <iostream>
using namespace std;

#define N 10

template <class T>

T min(T *num,int n)
{
T k = *num;
for (int i = 1;i < n;i ++) {
num ++;
if (*num < k)
k = *num;
};
return k;
}

int main()
{
int n = N + 1;
while ( n > N) {
cout << "Enter the length of your array (don't larger than 10):";
cin >> n;
}
char a1[N];
int a2[N];
float a3[N];
long a4[N];
for (int i = 0;i < n;i ++) {
cout << "a1[" << i + 1 << "] = ";
cin >> a1[i];
cout << "a2[" << i + 1 << "] = ";
cin >> a2[i];
cout << "a3[" << i + 1 << "] = ";
cin >> a3[i];
cout << "a4[" << i + 1 << "] = ";
cin >> a4[i];
}
cout << "Min in a1 is " << min(a1,n) << endl;
cout << "Min in a2 is " << min(a2,n) << endl;
cout << "Min in a3 is " << min(a3,n) << endl;
cout << "Min in a4 is " << min(a4,n) << endl;
return 0;
}

Answer 2

#include <iostream>

using namespace std;

template <class T>

T min(T *num,int n)
{
T k = *num;
for (int i = 1;i < n;i ++)
{
num ++;
if (*num < k)
k = *num;
};
return k;
}

int main()
{
int n;
cin >> n;
char a1[n];
int a2[n];
float a3[n];
long a4[n];
for (int i = 0;i < n;i ++)
cin >> a1[i] >> a2[i] >> a3[i] >> a4[i];
cout << min(a1,n) << endl << min (a2,n) << endl << min(a3,n) << endl << min(a4,n) << endl;
return 0;
}

Answer 3

恩 就正常重载就行啊