+ 1

# Assignment Operators problem...

I can't understand this: x *= 3; // equivalent to x = x * 3 x /= 2; // equivalent to x = x / 2 x %= 4; // equivalent to x = x % 4 so "X" is a number for example 2, so the first one would be 2*=3 //equivalent to 2=2*3 right? but where's the point of doing that, im so confused please help!

7th Apr 2017, 9:32 PM
Francesco Ranchi
+ 16
When we write something like x = x+1, Think it as, new x = old x + 1 So, x = 2+1 We are supposed to calculate/ assign the left hand side operand based on the right hand side formula.
7th Apr 2017, 9:42 PM
Shamima Yasmin
+ 11
Don't see it the mathematical way... How can I explain it? I'll try this: Imagine the variable is a box, named x (ok.... I know... maybe somehow it is signed as x). In your example you first put the number 2 in it: x=2; and then you take the number 2 out of it again multiply it by 3 and put the result 6 back in the box again. x=x*3; You can go to that box and see what's in there any time you need it. x*=3; is just a short way to write x=x*3;
7th Apr 2017, 9:45 PM
Tashi N
+ 6
for example: assuming we have the variable x which is assigned the value of 3 if we got x *= 3 then we do not have the situation of 3 = 3 * 3 instead we have it like this: x = 3*3 this seems a little bit out of world and pretty strange, but you gotta see it like that: variable x just points to a value with every new operation to the variable, not the variable itself is changed, but the so thought variable-arrow that points to a specific value so to say one could understand it like this more easily: x *= 3 --> x(new) = x(old) * 3 which is in this case: --> x(new) = 3 * 3 so now: x = 9 doing that again with the new value: --> x(new) = x(old) * 3 which is in this case: --> x(new) = 9 * 3 so now: x = 27 I hope I helped you with thatđ
7th Apr 2017, 9:56 PM
H3LL0FR14ND
+ 1
X's new value = X's old value * 3
7th Apr 2017, 9:39 PM
Dennis
+ 1
x Ă·= 5
7th Apr 2017, 10:22 PM
Edward