Lionheart
Exactly what does & mean by opportunity? I’m sure 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 number. 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 position that is last 1 and 1 equals 1.
Lionheart
Do you need more explanation – or can you simply instead skip it.
Do you run into this in just one of ACES guages and wished to understand how it worked?
Think about it you need to have counted in binary as a young child
Zero one ten eleven one hundred a hundred and something a hundred and ten a hundred and eleven.
I want to explain or even to you.
No No make him stop. We’ll talk, We’ll talk
FelixFFDS
Ron – I may have understood exactly exactly exactly what the AND operator implied – a time that is long – in university.
So utilizing your example, 3,5 OR gives me personally “6”?
N4gix
Hey dudes, exactly what does & suggest by possibility? I’m sure that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically correct, ” i am let’s assume that you just desired to understand the following:
& is simply the way that is”full of composing the “&” expression.
. Exactly like >: may be the way that is”full of composing “”.
(Hint: the expression is named an “ampersand” or “amp” for short! )
In FS XML syntax, its utilized such as this:
&& is similar as && is equivalent to and
I recently explained this in another post about a week ago.
You did XOR – exclusive OR
You compare the bits vertically – within my examples
You can get the image.
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 very last 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 final stack entry — (unary operator): decrements last stack entry ++ (unary operator): increments last stack entry
(binary operator): ”” offers 1 if last stack entry is more than forelast stack entry (binary operator): ” >=; (binary operator): ”=” gives 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 final stack entries are 1 offers 1 otherwise 0 || (binary operator): logical OR, if an individual for the 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”, if the final entry is 1, the forelast entry is employed, else the fore-forelast ( or perhaps one 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 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.: loads value from storage space and places along 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 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 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 (remember 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 in the brackets is performed, else the rule into the second pair of brackets ( just take also care to where SPACEs are allowed and where perhaps perhaps not) stop actually leaves the execution straight away, last stack entry is employed 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 down the stack, 7 states exactly just exactly how much entries, on the basis of the results of (A: Flaps handle index, quantity) ”case” extracts one of many seven figures. If (A: Flaps handle index, quantity) is 0 – 0, 1-1, 2-5. 6-30.