C#•3mo ago

✅ Addition with integers and floating point numbers

namespace Calculator.Model
{
    // The class for Addition "+".
    public class Addition
    {
        public static int Add(int number1, int number2)
        {

        }
    }
}

namespace Calculator.Model
{
    // The class for Addition "+".
    public class Addition
    {
        public static int Add(int number1, int number2)
        {

        }
    }
}

If i were to create this method, and someone would use a floating point number instead of an int. Would this method break? And if so, is there a way around this?

31 Replies

Salman•3mo ago

I think you could use float in that case , it would be able to accept both int and float inputs right now it'd be small -> large type conversion which will cause error but if you change tha params type to float and you get an int as a parameter then it'll be large -> small type convesion which is ffine

Merineth 🇸🇪OP•3mo ago

Neat So i just have the two inputs be float and it will implicity convert from an integer if it's input?

Salman•3mo ago

yes also change the return type to be float

MODiX•3mo ago

Salman

REPL Result: Success

public static float Add(float number1, float number2)
        {
return number1 + number2;
        }
Console.Write(Add(1,2));

public static float Add(float number1, float number2)
        {
return number1 + number2;
        }
Console.Write(Add(1,2));

Console Output

Compile: 443.149ms | Execution: 25.086ms | React with ❌ to remove this embed.

Salman•3mo ago

based on your usecase you might also want to use double or decimal etc

Merineth 🇸🇪OP•3mo ago

yeah i looked into that but float will be fine wont expect anything longer than 3 decimals long Also do you know how i can use the bot to try out some code? For example.

namespace Calculator.Model
{
    internal class Modulo
    {
        public static float Mod(float number1, float number2)
        {
            return number1 % number2;
        }
    }
}

namespace Calculator.Model
{
    internal class Modulo
    {
        public static float Mod(float number1, float number2)
        {
            return number1 % number2;
        }
    }
}

Does this mod work ?

Salman•3mo ago

to run the code using bot you need to run the bot command in this way :

!eval //your c# code here

!eval //your c# code here

just remove the namespaces and classes and paste the function and it's call as I did above

Merineth 🇸🇪OP•3mo ago

Ah Yay seem to work

Salman•3mo ago

you can spam the bot in #bot-spam though

Merineth 🇸🇪OP•3mo ago

Yeah i'll keep that in mind c: Just making sure it works

Salman•3mo ago

we've a dedicated channel for the bot xD

Merineth 🇸🇪OP•3mo ago

:catderp: But thanks! everything seems to work fine now c: happy friday

FusedQyou•3mo ago

A floating point is casted to an integer, removing the decimals that exist If you want this to work for floats, either make one for float with voerloading (like you did), or use INumber INumber is a rather new interface that combines common operations, including additions So instead of making multiple methods that practically do the same, you can put them under a single INumber So this is wrong. It doesn't accept floats, it just implicitly casts a float to an integer which gets rid of the decimals So if you were to add 1.5 to 1.5, it returns 2 instead of 3 Kind of hard to explain INumber, but here is a method that uses it (INumber implements IAdditionOperators)

public static T AddAll<T, TOther>(T t, List<TOther> c) where T : IAdditionOperators<T, TOther, T>
{
    foreach (var item in c)
    {
        t = t + item;
    }

    return t;
}

public static T AddAll<T, TOther>(T t, List<TOther> c) where T : IAdditionOperators<T, TOther, T>
{
    foreach (var item in c)
    {
        t = t + item;
    }

    return t;
}

But if you are a beginner, forget this. Just write overloads If you make a method of the same name, but different parameters, we call those overloads

Salman•3mo ago

I think it works just fine, as a beginner they don't need to implement those complex interfaces

FusedQyou•3mo ago

Wasn't my point And it doesn't work at all

Salman•3mo ago

how ?

FusedQyou•3mo ago

You don't need to use INumber here, you can use overload methods fine I already explained this

MODiX•3mo ago

FusedQyou

So if you were to add 1.5 to 1.5, it returns 2 instead of 3

Quoted by

<@191282187016863744> from #✅ Addition with integers and floating point numbers (click here)

React with ❌ to remove this embed.

MODiX•3mo ago

FusedQyou

REPL Result: Success

(int)1.5 + (int)1.5

(int)1.5 + (int)1.5

Result: int

Compile: 198.583ms | Execution: 15.108ms | React with ❌ to remove this embed.

MODiX•3mo ago

Salman

REPL Result: Success

public static float Add(float number1, float number2)
        {
return number1 + number2;
        }
Console.Write(Add(1.5f,1.5f));

public static float Add(float number1, float number2)
        {
return number1 + number2;
        }
Console.Write(Add(1.5f,1.5f));

Console Output

Compile: 460.299ms | Execution: 25.074ms | React with ❌ to remove this embed.

Salman•3mo ago

Works just fine it's returning 3 as it should, the function would work fine with both normal integers and floats without making any overloads or such

FusedQyou•3mo ago

Oh, the point is using float as a general method? That's not any better to be honest 😄

Salman•3mo ago

their problem was that they want to accept both normal ints and floating points so the above function solves that

FusedQyou•3mo ago

This is bad advice, don't do this to beginners This works fine with small numbers, but the whole point of the float type is to store a base and an exponent, and this will cause inaccuracies as with any type that contains decimals Your integer type is precise, and will never have this issue And this is why you don't give advice like this, especially since there is a better solution right in front of you with overloading

Salman•3mo ago

then can't double or decimal solve this problem with bigger numbers ?

FusedQyou•3mo ago

@Merineth Please use overloading instead of one method.

namespace Calculator.Model
{
    // The class for Addition "+".
    public class Addition
    {
        public static int Add(int number1, int number2)
        {
            return number1 + number2;
        }

        public static float Add(float number1, float number2)
        {
            return number1 + number2;
        }
    }
}

namespace Calculator.Model
{
    // The class for Addition "+".
    public class Addition
    {
        public static int Add(int number1, int number2)
        {
            return number1 + number2;
        }

        public static float Add(float number1, float number2)
        {
            return number1 + number2;
        }
    }
}

Salman•3mo ago

can you provide an example of how the above function with even decimal , double or float will not be able to handle the bigger numbers and will break the application ?

FusedQyou•3mo ago

Look up floating point arithmetic imprecision

Salman•3mo ago

I think that breaking the application is an exaggeration also you've missed the context as well : https://discord.com/channels/143867839282020352/1286653567150587916/1286656203211935766

MODiX•3mo ago

Merineth

wont expect anything longer than 3 decimals long

Quoted by

<@923267140960796713> from #✅ Addition with integers and floating point numbers (click here)

React with ❌ to remove this embed.

FusedQyou•3mo ago

I don't see a point in discussing this if you refuse to take the suggestions from the other party My point is clear, and I ask you don't give bad advice to beginners like this The correct way is either overloading or the usage of INumber(base) Even if this user were to expect a maximum size in their numbers, you still don't advice this They don't know about explicit/implicit conversions that happen here, and if they take advices like this and make an application with it they end up with a poorly build application that does not work. And if this worked for you, great, but let's not give this bad advice to beginners and give them the correct way as soon as possible

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✅ Addition with integers and floating point numbers