Exactly what does & suggest by opportunity? I am aware that & means ‘and’, but amp has wondering.
Where 3 5 & offers 1
The bits in each place in the 1st quantity (chr) must match bits in each place into 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 position that is last 1 and 1 equals 1.
Do you want more explanation – or could you simply instead skip it.
Do you run into this in another of ACES guages and desired to understand how it worked?
Think about it you really need to have counted in binary as a youngster
Zero one ten eleven a yubo lunch box hundred a hundred plus one a hundred and ten a hundred and eleven.
Allow me to explain or even you.
No No make him stop. We’ll talk, We’ll talk
Ron – i might have understood just what the AND operator suggested – a very long time ago – in university.
Therefore utilizing your instance, 3,5 OR gives me personally “6”?
Hey dudes, So what does & suggest by opportunity? I understand that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically proper, ” i am let’s assume that you just wished to understand the following:
& is simply the way that is”full of composing the “&” expression.
. Exactly like >: could be the “full method” of composing “”.
(Hint: the icon is named an “ampersand” or “amp” for short! )
In FS XML syntax, it really is utilized similar to this:
&& is the identical as && is equivalent to and
I simply explained this in another post about an ago week.
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
+ (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 very last two stack entries percent (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 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 higher than or add up to forelast stack entry <=; (binary operator): ” == (binary operator): gives 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 a person associated with final stack entries is 1 outcome is 1 otherwise 0! (unary operator): rational NOT, 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 the other way round. Check it out, notice 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 towards the right
D: duplicates final stack entry r: swaps final two stack entries s0, s1, s2.: shops final stack entry in storage for later use sp0, sp1, sp2.: (presumably) exactly the same as above l0, l1, l2.: lots value from storage space and places along 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: doesn’t work too dependable) dgrd (unary operator): converts levels to radians (also rddg available? ) pi: places p over 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
Others: if if last stack entry is 1, the rule in the brackets is performed (note that there’s no 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 when you look at the 2nd pair of brackets ( simply take also care to where SPACEs are permitted and where maybe maybe not) stop simply leaves the execution straight away, final stack entry is employed for further purposes instance difficult to explain, consequently a good example:
30 25 20 10 5 1 0 7 (A: Flaps handle index, quantity) instance
The figures 30 25 20 10 5 1 0 are pushed down the stack, 7 claims exactly just just how much entries, in line with the consequence of (A: Flaps handle index, quantity) ”case” extracts one of many seven numbers. If (A: Flaps handle index, quantity) is 0 – 0, 1-1, 2-5. 6-30.