CODEWITHSUNDEEP
c#.net
Leo Vo
Leo Vo

Reputation: 10330

Changing an Actual Parameter in C#

In C++, I have a function:

void MyFunction(int p)
{
    p=5;
}

Assume, I have:

int x = 10;
MyFunction(x); // x = 10
MyFunction(&x); // x = 5

How to archieve this in C# with condition: I create only one MyFunction.

Upvotes: 0

Views: 154

Answers (5)

A9S6
A9S6

Reputation: 6685

The point is that a lot of people think that Reference types are passed by reference and Value types are passed By Value. This is the case from a user's perspective, internally both Reference and Value types are passed By Value only. When a Reference type is passed as a parameter, its value, which is a reference to the actual object is passed. In case of Value types, their value is the value itself (e.g. 5).

StringBuilder sb = new StringBuilder();
SetNull(sb);
SetNull(ref sb);

if SetNull(...) sets the parameter to null, then the second call will set the passed in StringBuilder parameter to null.

Upvotes: 0

Oscar Mederos
Oscar Mederos

Reputation: 29863

You need to pass the parameter as reference. If you don't specify it, it automatically creates a copy to work inside the parameter instead of using the same reference.

How to do that? Just specify with the 'ref' word in method declaration:

void MyFunction(ref int p)
{
    p=5;
}

int x = 10;
MyFunction(ref x); // x = 5

Upvotes: 0

EMP
EMP

Reputation: 62031

In C# you would need to declare the method with a ref parameter, like this:

void MyFunction(ref int p)
{
    p=5;
}

If you then call it as MyFunction(ref x) the value of x in the caller will be modified. If you don't want it to be modified simply copy it to a dummy variable. You could create an overload of MyFunction that does this internally:

void MyFunction(int p)
{
    MyFunction(ref p);
}

It would technically not be "one function", as you want, but the code wouldn't be duplicated and to any human reading your code it would appear as one - but to the compiler it's two. You would call them like this:

int x = 10;
MyFunction(x); // x = 10
MyFunction(ref x); // x = 5

Upvotes: 2

Malcolm Post
Malcolm Post

Reputation: 515

C# does not have the equivalent functionality. If you declare the method to have a ref parameter, then you must also specify that the parameter is ref type when you call the method.

Upvotes: 1

John Saunders
John Saunders

Reputation: 161831

Your C++ function doesn't work the way you think it does. In fact, your code will not compile.

In C#, you would use the ref or out keywords:

void MyFunction1(out int p)
{
    p = 5;
}

void MyFunction2(ref int p)
{
    p = p + 1;
}

int x;
MyFunction1(out x); // x == 5
MyFunction2(ref x); // x == 6

Upvotes: 3

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