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Index › Programming Questions & Help › Syntax Questions › println specific number
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println specific number (Read 702 times)
clausneergaard
println specific number
Dec 7th, 2009, 3:26am  
hi, i'm wondering how i can get to print only specific numbers, for example

if i = *.0 then println(i)

... meaning only print i, if i is a 1.0, 2.0, 3.0 etc. don't print i if 1.2, 3.5 etc. can't seem to figure out why the below code doesn't work

Code:

void draw() {
  for (float i = 0; i < 100.0; i += 0.1) {  
    if (i == round(i)) {
      println(i);
    }
  }
}


thanks in advance
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blindfish
Re: println specific number
Reply #1 - Dec 7th, 2009, 3:45am  
Have you tried println(i) regardless (i.e. not in a condition).  This might answer your question:

Code:
void setup() {
  size(100,100);
  for (float i = 0; i < 100.0; i += 0.1) {  
    println(i);
    if (i%1 == 0) {
      println("round number: " + i);
    }
  }
}


Search for 'floating point inacuracy' to find the cause of the problem.  One solution is to work with whole numbers and then divide by 10/100/100 depending on how many decimal places you want...

Note also that using modulo is another way to do this test.
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Where is the what if the what is in why?
 
blindfish
Re: println specific number
Reply #2 - Dec 7th, 2009, 3:50am  
For example:

Code:
void setup() {
  size(100,100);
  for (float i = 0; i < 1000.0; i += 1) {  
    float number = i/10;
    if (number%1 == 0) {
	println("round number: ");
    }
    println(number);
  }
}
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Where is the what if the what is in why?
 
clausneergaard
Re: println specific number
Reply #3 - Dec 7th, 2009, 4:00am  
thanks blindfish, using modulo works great. and it makes sense. exactly the solution i was looking for.

cheers!
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