C#•13mo ago

understanding %

36 Replies

・OP•13mo ago

I don't understand why this is 1 as output

・OP•13mo ago

Its int so it should be a 179 or a 134 not a 23.1 or a 13.4

・OP•13mo ago

I remade it with double but still 1

Angius•13mo ago

It's the remainder

XE SparkyCracked•13mo ago

I respect the coding on mobile situation. Holy

・OP•13mo ago

I divide 100 by 3 and the opposite of the anwser, right?

Angius•13mo ago

13 % 5 = 3 Because 5 fits entirely inside of 13 twice 2 x 5 is 10 13 - 10 is 3

XE SparkyCracked•13mo ago

The coding term for '%' is Modulo

・OP•13mo ago

So i search for the closest multiplication and add numbers to it until its 100?

XE SparkyCracked•13mo ago

If you search up this definition it returns the following:

Modulo is a mathematical operation that returns the remainder of a division of two arguments. It is available in every programming language.

Modulo is a mathematical operation that returns the remainder of a division of two arguments. It is available in every programming language.

・OP•13mo ago

So like this?

・OP•13mo ago

So the modulo is 1 Alias %

XE SparkyCracked•13mo ago

Modulo is the remainder When dividing

Angius•13mo ago

Green is 13 Red and blue are both 5 The empty space, where another 5 cannot fit, is the remainder

XE SparkyCracked•13mo ago

so 100/33 = 3 if you take whats left: 100 - (33*3) = modulo In this case 1

・OP•13mo ago

So if i get this again i can just 100/3=33.3333 100/33=3 100 - (33*3)

XE SparkyCracked•13mo ago

Note we deal with whole numbers here

・OP•13mo ago

Yes, not doubles

XE SparkyCracked•13mo ago

Yeah

・OP•13mo ago

I think i understand it now, thank you everyone

XE SparkyCracked•13mo ago

100/3 = 33 100 - (333) = modulo of 100/3 1 100/33 = 3 100 - (333) = modulo of 100/33 1 I'll send the sudo code now:

c#
num1/intDevider = ans // whole number
num1 - (intDevider*ans) = modulo // your remainder as a whole number

c#
num1/intDevider = ans // whole number
num1 - (intDevider*ans) = modulo // your remainder as a whole number

Hope it makes sense @Ultra Giga Beginner You don't need to / twice ^^

・OP•13mo ago

It makes, but i don't know what a sudo code is yet

XE SparkyCracked•13mo ago

It's spelt 'pseudocode' but I get lazy lol. I describe it as english code...it doesn't run anywhere, but has some syntax so you can put it in code later. Used for discussing logic. Maybe @ZZZZZZZZZZZZZZZZZZZZZZZZZ has a better way to describe it

Angius•13mo ago

Well, if you wanted to write your own modulo method...

public static int Mod(int num, int divisor)
{
    var result = num / divisor;
    return num - divisor * result;
}

public static int Mod(int num, int divisor)
{
    var result = num / divisor;
    return num - divisor * result;
}

MODiX•13mo ago

Angius

REPL Result: Success

int Mod(int num, int divisor)
{
    var result = num / divisor;
    return num - divisor * result;
}

var myMod = Mod(14, 8);
var mod = 14 % 8;

new {
    MyMod = myMod,
    Mod = mod,
    Equal = myMod == mod
}

int Mod(int num, int divisor)
{
    var result = num / divisor;
    return num - divisor * result;
}

var myMod = Mod(14, 8);
var mod = 14 % 8;

new {
    MyMod = myMod,
    Mod = mod,
    Equal = myMod == mod
}

Result: <>f__AnonymousType0#1<int, int, bool>

{
  "myMod": 6,
  "mod": 6,
  "equal": true
}

{
  "myMod": 6,
  "mod": 6,
  "equal": true
}

Compile: 330.227ms | Execution: 65.915ms | React with ❌ to remove this embed.

Angius•13mo ago

Man, I'm glad we learned division with remainder in primary school, before we even learned proper division lol It seems to stump so many people

XE SparkyCracked•13mo ago

Yep

・OP•13mo ago

Okay, i can't comprehend this yet, but later i hope I'll. I kind of understand it but if i would need to write this out from head with logic i would not be able to

Angius•13mo ago

Exercise

XE SparkyCracked•13mo ago

Normally the case when learning new things in code It's normal don't stress. Just keep practicing

Angius•13mo ago

Try calculating some modulos, then check with C# if you got the answer right

Tvde1•13mo ago

÷ and % are not the same unfortunately

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understanding %

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