So what does & suggest by opportunity? I’m sure that & means ‘and’, but amp has wondering.
Where 3 5 & gives 1
The bits in each place in the 1st quantity (chr) must match bits in each place within the number that is second. Right right right Here just the people in red.
One other place either have 0 and 0 equals 0 or 1 and 0 equals 0. However the final place has 1 and 1 equals 1.
Do you need more explanation – or can you simply rather skip it.
Did you run into this in just one of ACES guages and desired to discover how it worked?
Think about it you really must have counted in binary as a young child
Zero one ten eleven a hundred a hundred and another 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 may have understood exactly what the AND operator suggested – a number of years ago – in university.
Therefore making use of your instance, 3,5 OR gives me personally “6”?
Hey dudes, What does & suggest by possibility? 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 desired to understand the following:
& is only the “full means” of composing the “&” icon.
. Exactly like >: could be the “full method” of writing “”.
(Hint: the sign is known as an “ampersand” or “amp” for short! )
In FS XML syntax, it really is utilized like this:
&& is the identical as && is equivalent bristlr mobile site to and
I simply explained this in another post of a week ago.
You did XOR – exclusive OR
You compare the bits vertically – in my own 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
+ (binary operator): adds the very last two stack entries – (binary operator): subtracts the past two stack entries * (binary operator): multiplies the past two stack entries / (binary operator): divides the very last two stack entries percent (binary operator): rest divides the past two stack entries /-/ (unary operator): reverses indication of final stack entry — (unary operator): decrements last stack entry ++ (unary operator): increments final stack entry
(binary operator): ”” provides 1 if final stack entry is more than forelast stack entry (binary operator): ” >=; (binary operator): ”=” gives 1 if final stack entry is higher than or add up to forelast stack entry <=; (binary operator): ” == (binary operator): offers 1 if both final final stack entries are equal && (binary operator): ”&&” logical AND, if both final 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 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 is used, else the fore-forelast ( or even 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 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 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 on 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 executed (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 within the brackets is performed, else the rule within the 2nd pair of brackets ( simply take also care to where SPACEs are permitted and where maybe maybe not) stop renders the execution straight away, final stack entry can be used for further purposes case difficult to explain, consequently a good example:
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 down the stack, 7 claims just exactly exactly how much entries, on the basis of the outcome 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.