Problem with adding parameters

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Hello!

I am new to programming and am experiencing issues with a the following exercise:

class program
{
public static void Main(string[] args)
{
string n=Calculator.WriteNumber(53+28);
Console.WriteLine(n);

Console.ReadLine();

I am confused by the (53 +28) part.How do I add an arithmetic sign?

What I have tried:

I created the following class which compiles( when Calculator.WriteNumber(53,28);)

class Calculator
{

public static string WriteNumber(int a ,int b)
{
int c = a + b;
string result = c.ToString();
return result;
}

Posted 10-Feb-18 4:34am

Member 13671506

Updated 10-Feb-18 5:32am

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4 solutions

Solution 1

53+28 counts as one parameter (so you still have to supply a second one): when you call WriteNumber, it will take 81 (i.e. 53 + 28) as parameter a. It does not get differently treated because there is an arithmetic sign: 53 + 28 evaluates to 81 which is one int and here you pass it as parameter a.

Posted 10-Feb-18 4:40am

Thomas Daniels

Comments

Member 13671506 10-Feb-18 10:47am

Thank you for your reply. If it is treated as one parameter , how do I add a second one without changing the look of string n=Calculator.WriteNumber(53+28)?

Dave Kreskowiak 10-Feb-18 10:50am

Because your WriteNumber method takes two parameters, you have to supply the parameters with a comman between them. Are you looking for:

    string n = Calculator.WriteNumber(53, 28);

Maciej Los 10-Feb-18 10:48am

Well explained.

Thomas Daniels 10-Feb-18 10:50am

Thank you!

Member 13671506 10-Feb-18 10:48am

I can say string n= Calculator.WriteNumber(53+87,0) but I don't think it will be accepted.

Thomas Daniels 10-Feb-18 10:50am

Yes, it would be accepted. 53 + 87 (so actually 140) is the first parameter 'a', 0 is the second parameter 'b'.

It will be accepted because WriteNumber wants two ints. And you have two ints: 140 (53 + 87) and 0.

Solution 2

You could try

C#

string n = Calculator.WriteNumber(53, 28);

instead.
The addition is realized by the WriteNumber method.
You can perform operations on method's parameters prior to passing them:

C#

string n = Calculator.WriteNumber(10 - 5, 2 * 2);

This is equivalent to

C#

string n = Calculator.WriteNumber(5, 4);

and will return 9.

Posted 10-Feb-18 4:53am

phil.o

Comments

Member 13671506 10-Feb-18 10:56am

The solution has to be in the format string n=Calculator.WriteNumber(52 + 28); and the result has to be 81. I submitted (53,28) but my teacher said that It has to be (52 + 28)

phil.o 10-Feb-18 11:06am

Then you have to define your method with only one parameter, but then there is no point to have a method here since there is nothing to compute (you already computed the result while calling the method).
It is unclear what you (or your teacher) really expect:
- either you keep two arguments for your method but then you have to call the method with two arguments.
- or you give just one parameter to the method and call it with this single parameter, but as I said there is no point anymore to have a method then (since there is nothing more to compute).

Solution 3

To pass two parameters, you separate them with a comma, just as you do when you declare the method:

string n=Calculator.WriteNumber(53, 28);

If you want to pass an operator, then I'd do it with an enum:

C#

public enum Operators
   {
   Plus,
   Minus,
   Times,
   DivideBy,
   }

Then:

public static string Calculate(int a, Operators op, int b)
    {
    int result;
    switch (op)
        {
        case Operators.Plus: result = a + b; break;
        case Operators.Minus: result = a - b; break;
        case Operators.Times: result = a * b; break;
        case Operators.DivideBy: result = a / b; break;
        default: throw new ArgumentException("Unknown Operator: " + op);
        }
    return result.ToString();
    }

You then call it like this:

string n = Calculator.Calculate(53, Operators.Plus, 28);

Note that I changed the name of your method: don't call something "WriteNumber" unless it actually writes something!

Posted 10-Feb-18 4:54am

OriginalGriff

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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Member 13671506 · Accepted Answer · 2018-02-10T04:59:00

Solution 4

public static string WriteNumber(int a ,int b=0)
string n=Calculator.WriteNumber(53+28);

The point of the exercise was for me to use default value and optional parameters.

This is the solution, thank you all for your help!

Posted 10-Feb-18 4:59am

Member 13671506

Updated 10-Feb-18 5:22am

v3

Comments

PIEBALDconsult 10-Feb-18 13:51pm

Please don't answer your own question.

Member 13671506 10-Feb-18 14:20pm

OK

Problem with adding parameters

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