C ++中的原型

Question

If I prototype a function above the main function in my code, do I have to include all parameters which have to be given? Is there a way how I can just prototype only the function, to save time, space and memory?

Here is the code where I came up with this question:

#include <iostream>

using namespace std;

int allesinsekunden(int, int, int);

int main(){
    int stunden, minuten, sekunden;

    cout << "Stunden? \n";
    cin >> stunden;
    cout << "Minuten? \n";
    cin >> minuten;
    cout << "Sekunden= \n";
    cin >> sekunden;

    cout << "Alles in Sekunden= " << allesinsekunden(stunden, minuten, sekunden) << endl;
}

int allesinsekunden (int h, int m, int s) {
    int sec;

    sec=h*3600 + m*60 + s;

    return sec;

}

Answer 1

"If I prototype a function above the main function in my code, do I have to include all parameters which have to be given?"

Yes, otherwise the compiler doesn't know how your function is allowed to be called.
Functions can be overloaded in c++, which means functions with the same name may have different number and type of parameters. Such the name alone isn't distinct enough.

"Is there a way how I can just prototype only the function, to save time, space and memory?"

No. Why do you think it would save any memory?

Answer 2

No, because it would add ambiguity. In C++ it's perfectly possible to have two completely different functions which differ only in the number and/or type of input arguments. (Of course, in a well-written program what these functions do should be related.) So you could have

int allesinsekunden(int, int, int)
{
//...
}

and

int allesinsekunden(int, int)
{
//...
}

If you tried to 'prototype' (declare) one of these with

int allesinsekunden;

how would the compiler know which function was being declared? Specifically how would it be able to find the right definition for use in main ?

Answer 3

您必须声明函数的完整签名，即名称，返回值，带有类型的所有参数，它们的常数等。