So What Does & Suggest?
Lionheart
So what does & suggest by possibility? I understand that & means ‘and’, but amp has wondering.
Where 3 5 & offers 1
The bits in each place in the first quantity (chr) must match bits in each place within the 2nd quantity. Right right right Here just the people in red.
One other position either have actually 0 and 0 equals 0 or 1 and 0 equals 0. Nevertheless the final place has 1 and 1 equals 1.
Lionheart
Do you need more explanation – or could you simply instead skip it.
Do you run into this in another of ACES guages and desired to know how it worked?
Think about it you have to have counted in binary as a young child
Zero one ten eleven a hundred a hundred and another one hundred and ten a hundred and eleven.
Allow me to explain or even to you.
No No make him stop. We’ll talk, We’ll talk
FelixFFDS
Ron – i might have understood exactly just what the AND operator implied – a time that is long – in university.
So making use of your instance, 3,5 OR gives me personally “6″?
N4gix
Hey dudes, just what does & mean by opportunity? I am aware that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically proper, ” i am let’s assume that you just desired to understand the following:
& is only the way that is”full of writing the “&” sign.
. Exactly like >: could be the way that is”full of composing “”.
(Hint: the sign is known as an “ampersand” or “amp” for short! )
In FS XML syntax, its utilized such as this:
&& is the identical as && is equivalent to and
I simply explained this in another post of a week ago.
You did XOR – exclusive OR
You compare the bits vertically – within my examples
The picture is got by you.
A 1 OR 0 is 1 A 0 OR 1 is 1 A 1 OR 1 is 1 A 0 OR 0 is 0
A 1 OR 0 is 1 A 0 OR 1 is 1 A 1 OR 1 is 0
N4gix
+ (binary operator): adds the very last two stack entries – (binary operator): subtracts the past two stack entries * (binary operator): multiplies the final two stack entries / (binary operator): divides the past two stack entries per cent (binary operator): rest divides the final two stack entries /-/ (unary operator): reverses indication of final stack entry — (unary operator): decrements last stack entry ++ (unary operator): increments final stack entry
(binary operator): ”” offers 1 if final stack entry is higher than forelast stack entry (binary operator): ” >=; (binary operator): ”=” provides 1 if final stack entry is higher than or add up to forelast stack entry <=; (binary operator): ” == (binary operator): provides 1 if both final final stack entries are equal && (binary operator): ”&&” rational AND, if both final stack entries are 1 offers 1 otherwise 0 || (binary operator): logical OR, if one of this final stack entries is 1 result is 1 otherwise 0! (unary operator): rational never, toggles last stack entry from 1 to 0 or 0 to at least one? (ternary operator): ”short if-statement”, in the event that final entry is 1, the forelast entry can be used, else the fore-forelast ( or perhaps one other way round. Test it, view it)
& (binary operator): ”&” bitwise AND | (binary operator): bitwise OR
(unary operator): bitwise NOT, toggles all bits (binary operator): ” (binary operator): ”” change bits of forelast stack entry by last stack actions off to the right
D: duplicates last stack entry r: swaps final two stack entries s0, s1, s2.: shops final stack entry in storage for later use sp0, sp1, sp2.: (presumably) equivalent as above l0, l1, l2.: lots value from storage space and places together with stack
(unary operator): provides next smallest integer dnor (unary operator): normalizes degrees (all values are ”wrapped around the group” to 0°-360°) rnor (unary operator): normalizes radians (all values are ”wrapped across the circle” to 0-2p) (NOTE: does not work too reliable) dgrd (unary operator): converts levels to radians (also rddg available? ) pi: places p on the top of stack atg2 (binary operator): gives atan2 in radians (other trigonometric functions? Sin, cos, tg? Other functions? Sqrt, ln? ) max (binary operator): provides greater of final two stack entries min (binary operator): provides the smaller of final two stack entries
Other people: if if final stack entry is 1, the rule within the brackets is performed (remember that there isn’t any AREA between ”if” and ””) if els if final stack entry is 1, the rule within the brackets is performed, else the rule into the 2nd pair of brackets ( just simply simply take also care to where SPACEs are permitted and where perhaps maybe not) stop departs the execution immediately, final stack entry is employed for further purposes instance difficult to describe, consequently a good example:
30 25 20 10 5 1 0 7 (A: Flaps handle index, number) situation
The figures 30 25 20 10 5 1 0 are forced along the stack, 7 states exactly exactly how much entries, on the basis of the consequence of (A: Flaps handle index, quantity) ”case” extracts one of the seven figures. If (A: Flaps handle index, quantity) is 0 – 0, 1-1, 2-5. 6-30.