Lionheart
Exactly what does & suggest by opportunity? I understand that & means ‘and’, but amp has wondering.
Where 3 5 & offers 1
The bits in each position in the 1st quantity (chr) must match bits in each place into the 2nd quantity. 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 last place has 1 and 1 equals 1.
Lionheart
Do you need more explanation – or could you simply instead skip it.
Did you run into this in just one of ACES guages and desired to understand how it worked?
Think about it you really must have counted in binary as a young child
Zero one ten eleven one hundred a hundred and another 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
FelixFFDS
Ron – i might have understood exactly exactly what the AND operator intended – a time that is long – in university.
Therefore utilizing your instance, 3,5 OR gives me personally “6”?
N4gix
Hey dudes, just what does & suggest by opportunity? I’m sure that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically correct, ” i am let’s assume that you merely desired to understand the following:
& is simply the way that is”full of composing the “&” sign.
. Exactly like >: could be the way that is”full of composing “”.
(Hint: the expression is called 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
The bits are compared by you 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 ethiopianpersonals.com two stack entries – (binary operator): subtracts the final two stack entries * (binary operator): multiplies the final two stack entries / (binary operator): divides the very last two stack entries per cent (binary operator): rest divides the very last two stack entries /-/ (unary operator): reverses indication of last stack entry — (unary operator): decrements last stack entry ++ (unary operator): increments stack entry that is last
(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 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 final stack entries are 1 offers 1 otherwise 0 || (binary operator): logical OR, if a person of this final 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 one 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): ”” shift bits of forelast stack entry by final stack steps 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.: 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 across the group” to 0°-360°) rnor (unary operator): normalizes radians (all values are ”wrapped around the circle” 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’s absolutely no AREA between ”if” and ”<” but one after it and at least one SPACE before ”>”) if < . >els if last stack entry is 1, the code within the brackets is performed, else the rule into the second group of brackets ( just simply take also care to where SPACEs are permitted and where maybe maybe not) stop renders the execution instantly, final stack entry can be used 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 forced along the stack, 7 claims just exactly just how entries that are much 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.