如何实现通用的max函数? [英] How to implement generic max function?

问题描述

我知道这是因为模板函数的返回类型与第一个参数(T)的返回类型相同.
如何修改此模板，使其在所有情况下都能正常运行?

I know this is because the return type of the template function is same as that of the first argument (T).
How can I modify this template so that, it behaves correctly in all cases?

#include <iostream>
using namespace std;
template <typename T, typename U>
T max(T x, U y)
{
    return x>y ? x : y;
}

int main()
{
    cout<<max(17.9,17)<<"\n";
    cout<<max(17,17.9)<<"\n";
}

输出:

17.9  
17

推荐答案

输出正确.您从未指定类型，因此抱怨它没有使用您希望它使用的类型是不合理的.如果你想要一个特定的类型，你必须确保两个参数都是那个类型.

The output is correct. You never specified a type, so complaining that it didn't use the type you wanted it to use is not reasonable. If you want a specific type, you have to ensure both parameters are that type.

您可以将第一个参数转换为 double.或者你可以专门调用
max.如果您真的非常想要，您也可以专门化 max 和类似的组合.

You could cast the first parameter to a double. Or you could specifically invoke
max<double, double>. You could also specialize max<int, double> and similar combinations if you really, really wanted to.

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