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[Canned Turkey]

I know the rule, if the last digit is 0 or 5, but I need something that uses math,
i.e.

if x/5 = 1 or 2 or 3 or 4 or 5...
then
do the thing
else
don't do that thing

But, I don't want to check the specific number,
i.e.

if x = 5 or 10 or 15 or 20...
then
do the thing
else
don't do that thing

I just need a function that if the input is divisible by 5, than it returns true.
I'm using GML, but basic math should translate anyway.

[Glyph]

if ((x mod 5) == 0) ...

[Canned Turkey]

if ((x mod 5) == 0) ...

Thank you!

[Cheezmeister]

The "remainder" operation is called modulo, and it's usually present as either mod or "%" (percent) in just about every programming language ever. See http://en.wikipedia.org/wiki/Modular_arithmetic

[Canned Turkey]

The "remainder" operation is called modulo, and it's usually present as either mod or "%" (percent) in just about every programming language ever. See http://en.wikipedia.org/wiki/Modular_arithmetic

Thanks for letting me know, no wonder I was having a hard time finding anything other than hard math.

[Columbo]

I suppose you could also do something like:

if ((x/5)*5 == x)

If your % key was broken or something 

[Glyph]

Yeah, or even

if (abs(cos(x/5 * pi)) == 1) ...

 

[oahda]

Code:

for (int i = 0; true; i += 5) 
    if (x == i || x == -i)
        break;

[Rarykos]

Yeah, or even

if (abs(cos(x/5 * pi)) == 1) ...

 

Wow, this must be the most complicated way! I'm so writing this on my next job interview when I'm asked this question 

[DocProctopus]

Code:

for (int i = 0; true; i += 5) 
    if (x == i || x == -i)
        break;

This will work but it's very inefficient. A single modulo operation is the best way to do this.

[Marchal_Mig12]

Binary operations would be more efficient imo.

[Christian Knudsen]

Code:

for (int i = 0; true; i += 5) 
    if (x == i || x == -i)
        break;

This will work but it's very inefficient. A single modulo operation is the best way to do this.

[Sik]

Yeah, or even

if (abs(cos(x/5 * pi)) == 1) ...

 

...I hate you. So much.

[Gtoknu]

some go with human-thinking:

Code:

int lastDigit = x % 10;
if(lastDigit == 5 or lastDigit == 0)
{
    functionCall();
}

EDIT:

The "remainder" operation is called modulo, and it's usually present as either mod or "%" (percent) in just about every programming language ever. See http://en.wikipedia.org/wiki/Modular_arithmetic

By the way, I think this: http://en.wikipedia.org/wiki/Modulo_operation is far more informative for a beginner.

[alvarop]

Can anyone explain why anyone would use anything but modulo?

[Gtoknu]

Can anyone explain why anyone would use anything but modulo?

We wouldn't. But it's funny to point out other ways to solve a problem.

[Sik]

Can anyone explain why anyone would use anything but modulo?

Reason #1: you don't know it even exists (which seems to be the case here), although I have to admit it's usually one of the first things you're taught (alongside + - * /) because of how easy and useful it is.

Reason #2: your language doesn't have that operator (or you don't know if it does), you need to come up with a workaround in that case (I believe some Basic variants have this issue).

[Kyle Preston]

Yeah, or even

if (abs(cos(x/5 * pi)) == 1) ...

Just as an fyi, this will work with some languages/versions of languages but it's probably not a good practice to implement. As an example, this works with Python 2 point whatever, but in Python 3, fractions < 1 aren't rounded to 0 and hence will not give you the same as % 5.  

Just my two cents.

[oahda]

Yeah, or even

if (abs(cos(x/5 * pi)) == 1) ...

Just as an fyi, this will work with some languages/versions of languages but it's probably not a good practice to implement. As an example, this works with Python 2 point whatever, but in Python 3, fractions < 1 aren't rounded to 0 and hence will not give you the same as % 5.  

Just my two cents.

if (abs(cos(round(x/5) * pi)) == 1) ...

Just keep on adding those functions.

[Ads]

Another method:

Code:

double mod(double value, double m) {
  return value - m * Math.floor(value / m);
}

Test output:

Code:

-10 mod 3 = 2.0
-9 mod 3 = 0.0
-8 mod 3 = 1.0
-7 mod 3 = 2.0
-6 mod 3 = 0.0
-5 mod 3 = 1.0
-4 mod 3 = 2.0
-3 mod 3 = 0.0
-2 mod 3 = 1.0
-1 mod 3 = 2.0
0 mod 3 = 0.0
1 mod 3 = 1.0
2 mod 3 = 2.0
3 mod 3 = 0.0
4 mod 3 = 1.0
5 mod 3 = 2.0
6 mod 3 = 0.0
7 mod 3 = 1.0
8 mod 3 = 2.0
9 mod 3 = 0.0