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Start your free trialstella J
2,234 Pointsbetween --a and a--
I can understand how value 'a' has been changed but don't get how other values are changed.
First why is 'c' 9? it's statement says 'c = a ++'...... I could understand this when I think of c as original value a..
but what I really don't understand is the change of value d. when it was 'd = --a'. it is 9 while 'd = a--', it is 10. how? why?
help me out.
2 Answers
Mike Baxter
4,442 PointsOh, that's interesting, I'd never thought of that before! But it makes sense.
--a and a-- do the same operation to the variable a. They both subtract 1 from a.
The only difference is in how it affects something else, like this variable d (or whatever you want to call it, it's just a variable).
Now, when we call
d = --a;
the operation on a happens first, before the value of d gets assigned. So d takes on the modified value of a. (Not the original value).
Alternatively, when we call
d = a--;
the operation on a happens after d gets assigned whatever a was initially. That is to say, after d takes the initial value of a, only then do we subtract 1 from a.
Personally I'm going to try to memorize it by thinking like this, the d variable is going to eat up the a as soon as it can. If it can catch it before the operation happens to a , then it will take on the original value. But if the operation happens first, then d takes the modified value of a , NOT the original.
Hopefully that helps! Let me know if I can clarify anything.
David Tonge
Courses Plus Student 45,640 PointsI haven't taken the iOS course but I'm pretty sure the problem you're having with understanding this concept is similar to javascript. --a and a-- is kinda the same thing but the only difference is when echoed you'll get the value of the variable before the arithmetic operator is applied.
stella J
2,234 Pointsstella J
2,234 Pointsthank you ! it really helps me a lot : )
Mike Baxter
4,442 PointsMike Baxter
4,442 PointsGlad I could help, Haengun Chung!