Exactly what does & suggest by opportunity? I understand that & means ‘and’, but amp has wondering.
Where 3 5 & provides 1
The bits in each place in the 1st quantity (chr) must match bits in each place into the number that is second. Right Here just the ones in red.
One other place either have actually 0 and 0 equals 0 or 1 and 0 equals 0. Nevertheless the position that is last 1 and 1 equals 1.
Do you need more alua mobile explanation – or can you just instead skip it.
Do you run into this in just one of ACES guages and wished to discover how it worked?
Think about it you really need to have counted in binary as a youngster
Zero one ten eleven a hundred a hundred and something one hundred and ten a hundred and eleven.
I’d like to explain or even you.
No No make him stop. We’ll talk, We’ll talk
Ron – i might have understood just exactly what the AND operator designed – a very long time ago – in university.
Therefore utilizing your instance, 3,5 OR gives me personally “6”?
Hey guys, exactly what does & suggest by opportunity? I am aware that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically proper, ” I’m let’s assume that you merely wished to understand the following:
& is simply the way that is”full of composing the “&” icon.
. Just like >: may be the way that is”full of composing “”.
(Hint: the sign is known as an “ampersand” or “amp” for short! )
In FS XML syntax, it really is utilized such as this:
&& is the identical as && is equivalent to and
I recently explained this in another post of an ago week.
You did XOR – exclusive OR
You compare the bits vertically – during 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
+ (binary operator): adds the last two stack entries – (binary operator): subtracts the past two stack entries * (binary operator): multiplies the very last two stack entries / (binary operator): divides the past two stack entries percent (binary operator): rest divides the very last two stack entries /-/ (unary operator): reverses sign of final stack entry — (unary operator): decrements last stack entry ++ (unary operator): increments stack entry that is last
(binary operator): ”” provides 1 if final stack entry is more than forelast stack entry (binary operator): ” >=; (binary operator): ”=” provides 1 if final stack entry is more 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 provides 1 otherwise 0 || (binary operator): logical OR, if an individual associated with last stack entries is 1 outcome 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 is employed, else the fore-forelast ( or perhaps the 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 final stack actions off to the right
D: duplicates stack that is last r: swaps final two stack entries s0, s1, s2.: shops stack that is last in storage for later use sp0, sp1, sp2.: (presumably) exactly the same as above l0, l1, l2.: loads 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 around the group” to 0-2p) (NOTE: does not work too dependable) dgrd (unary operator): converts levels to radians (also rddg available? ) pi: places p at the top of stack atg2 (binary operator): gives atan2 in radians (other trigonometric functions? Sin, cos, tg? Other functions? Sqrt, ln? ) maximum (binary operator): provides the 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 code within the brackets is performed (remember that there isn’t any AREA between ”if” and ”<” but one after it and at least one SPACE before ”>”) if < . >els if final stack entry is 1, the rule within the brackets is performed, else the rule into the 2nd collection of brackets ( just take also care to where SPACEs are permitted and where perhaps perhaps not) stop simply leaves the execution instantly, last stack entry is employed for further purposes instance hard to explain, consequently an illustration:
30 25 20 10 5 1 0 7 (A: Flaps handle index, quantity) situation
The figures 30 25 20 10 5 1 0 are forced along the stack, 7 states just exactly just how entries that are much in line with the consequence of (A: Flaps handle index, quantity) ”case” extracts one of many seven figures. If (A: Flaps handle index, number) is 0 – 0, 1-1, 2-5. 6-30.