C＃输出参数，为什么我给我的变量赋值，仍然无法返回？

Question

I have a simple method like below,

        private int Increment(out n) {
            int n = 46;
            return n++;
        }

as my understanding, using out must initialise the variable first, but I still got the error of

return n++ 

The out parameter 'n' must be assigned to before control leaves the current method

I also got an error of my variable

int n = 46;

A local or parameter named 'n' cannot be declared in this scope because that name is used in an enclosing local scope to define a local or parameter

My question is, why I can't declare a new variable inside my method, it seems I must declare my variable outside of the method, then assigned the value inside the method.

If I just out, it means I must pass variable int to my method, can't I created inside the method? And out the parameter's pointer I declare inside my method?

Answer 1

The out declaration 'declares' the variable, you just have to set it in the method:

private int Increment(out int n)
{
    n = 46;
    return n++;
}

Note that this method returns 46, but n is 47.

Answer 2

You can write your method like this. It will return perfect output.

private int Increment(out int n)
{
    n = 46;
    n++;
    return n;
}

This will return 47 as output.

Answer 3

Not sure what you are trying to do but it's not the way it works. Look at the example (its not the simplest way to increment a number it's just a POC of how the ref and out parameters work).

static void Main(string[] args)
{
    int incrementingNumber = 0;

    while(true)
    {
        Increment(ref incrementingNumber, out incrementingNumber);

        Console.WriteLine(incrementingNumber);
        Thread.Sleep(1000);
    }
}

private static void Increment(ref int inNumber ,out int outNumber)
{
    outNumber = inNumber + 46;
}

the output is : 46 92 138 184 230 276 322 368 414 460 506 552

Answer 4

1. basically, you did not declare n correctly.
2. then in your function (line3), as n is declared already, you cannot declare it as int again.
3. your function will always return 46, because you are reassigning it in line 3 to 46
4. your function will return 46, but will set n to 47, im not sure if this is what you want.

If you want to just increment by 1 you could run this Code anywhere:

n++; // increment n by 1

n += 46; // increment n by 46

If you really want a function, I doubt that you need both, out int n and return n

this code will increment your variable n:

private static void Main()
{
    int n = 46;
    // some code here
    IncrementByOne(ref n);
    Console.WriteLine("N = " + n);
}    

/// this function will increment n by 1
private static void IncrementByOne(ref int n) { n++; }

This code will return an incremented integer

private static void Main ()
{
    int n = 46;
    // some code here
    n = IncrementByOne(n);
    Console.WriteLine("N = " + n);
}    

/// this function will return an integer that is incremented by 1
private static int IncrementByOne(int n)
{
    n++; //increment n by one
    return n;
}

If you want to declare n in your function, you have to declare it earlier

static void Main(string[] args)
{
    int n = 46; // you have to set n before incrementing
    // some code here
    IncrementByOne(out n);
    Console.WriteLine("N = " + n);
}
/// this function will always set variable n to 47
private static void IncrementBy47(out int n)
{
    n = 46; // assign n to 46
    n++; // increase n by one
}

This could also be shortened like:

/// This function will always set variable n to 47
private static void SetVarTo47(out int n) { n = 47; }

Hope that helps understanding what they do.