Little bit confused on the Mod Operator
Here's a sample from my review:
7) How many times does the letter A print out in the following segment of code?
A) 10 B) 12 C) 15 D) 20 E) NOTACode:int i=0; while (i < 20) { i++; if (i%5 > 2) { continue; } printf(“A”); }
The answer is b, but I dont understand the if statement.
For instance, I do understand if something like 12%5, which is 2 because 5 can only go up to 10, so the remainder is 2.
But when it's something like 19%200, it throws me off
If the remainder of i / 5 is > 2 then don't print an "A"
Mainframe assembler programmer by trade. C coder when I can.
1234567890
YYNNYYYNNY = 6
Now multiply that by 2.
Quzah.
Hope is the first step on the road to disappointment.
19 / 200 is zero, remainder 19.But when it's something like 19%200, it throws me off
Mainframe assembler programmer by trade. C coder when I can.
Or to be really picky, that segment of code wont compile (as posted) because it contains smart-quotes around the "A", rather than regular ones.
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Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
So if thats the case, then wouldnt the program read like this:Originally Posted by Dino
1=5Code:if (i%5 > 2)
2=5
3=5
4=5
5=0
6=1
7=2
8=3
9=4
10=0
so i= 1, 2, 3, 4, 8 and 9 is true since it is bigger than two
am I understanding this correctly?
No, look back at my earlier post:
1%5 = 1 > 2 = F
2%5 = 2 > 2 = F
3%5 = 3 > 2 = T
4%5 = 4 > 2 = T
5%5 = 0 > 2 = F
The % operator is basic division, before they taught you about decimals.
1 / 5 = 0 R 1
2 / 5 = 0 R 2
3 / 5 = 0 R 3
4 / 5 = 0 R 4
5 / 5 = 1 R 0
So you look at the remainder, and see if that is bigger than 2.
Quzah.
Hope is the first step on the road to disappointment.
No that's not quite right. It should be:
0≡0
1≡1
2≡2
3≡3
4≡4
5≡0
6≡1
7≡2
8≡3
9≡4
10≡0
All of those are modulo five of course. Note how the right is never 5 or greater.
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
1 % 5 = 1
2 % 5 = 2
3 % 5 = 3 First match
4 % 5 = 4. 2nd match
5 % 5 = 0
6 % 5 = 1
7 % 5 = 2
8 % 5 = 3 3rd match
and so on
and so on
Mainframe assembler programmer by trade. C coder when I can.
Thank you! I was actually looking the other way around
Thanks alot, that clears up
