Just 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 1st quantity (chr) must match bits in each place within the second quantity. Right right right Here just the people in red.
One other place either have actually 0 and 0 equals 0 or 1 and 0 equals 0. Nevertheless the final place has 1 and 1 equals 1.
Do you need more explanation – or can you simply instead skip it.
Did you encounter this in just one of ACES guages and wished to understand how it worked?
Come on you have to have counted in binary as a young child
Zero one ten eleven a hundred a hundred plus one a hundred and ten a hundred and eleven.
I would ike 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 time that is long – in university.
Therefore with 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 correct, ” i am let’s assume that you merely wished to understand the following:
& is only the way that is”full of composing the “&” sign.
. Exactly like >: could be the way that is”full of composing “”.
(Hint: the sign is named 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 simply explained this in another post of 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 very last two stack entries * (binary operator): multiplies the past two stack entries / (binary operator): divides the past two stack entries percent (binary operator): rest divides the final 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): ”” offers 1 if last stack entry is higher than forelast stack entry (binary operator): ” >=; (binary operator): ”=” provides 1 if last stack entry is more than or add up to forelast stack entry <=; (binary operator): ” == (binary operator): offers 1 if both final final stack entries are equal && (binary operator): ”&&” rational AND, if both last stack entries are 1 offers 1 otherwise 0 || (binary operator): logical OR, if one of this final stack entries is 1 outcome is 1 otherwise 0! (unary operator): rational never, delete beetalk account 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, 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 off to 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) 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 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 rule in the brackets is performed (remember that there is absolutely 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 in the brackets is performed, else the rule into the 2nd pair of brackets ( simply take also care to where SPACEs are permitted and where perhaps maybe not) stop actually leaves the execution straight away, final stack entry can be used for further purposes instance difficult 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 pressed along the stack, 7 claims just how much entries, on the basis of the results 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.