Just What Does & Suggest?
So what does & mean 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 2nd quantity. https://datingmentor.org/420-dating/ 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 want more explanation – or can you simply instead skip it.
Do you run into this in another of ACES guages and desired to discover how it worked?
Think about it you really need to have counted in binary as a young child
Zero one ten eleven a hundred a hundred and another one hundred and ten a hundred and eleven.
Allow me to explain or even to you.
No No make him stop. We’ll talk, We’ll talk
Ron – i might have understood exactly exactly what the AND operator implied – a time that is long – in university.
So making use of your instance, 3,5 OR gives me personally “6”?
Hey dudes, exactly what does & suggest by possibility? I’m sure that & means ‘and’, but amp has wondering. Many thanks,
While Ron is “technically proper, ” i am assuming that you just wished to understand the following:
& is only the way that is”full of composing the “&” sign.
. Exactly like >: could be the “full means” of composing “”.
(Hint: the icon is known as 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 – in my own 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 past two stack entries – (binary operator): subtracts the very last two stack entries * (binary operator): multiplies the very last two stack entries / (binary operator): divides the very last two stack entries percent (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 more 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): ”&&” logical AND, if both final stack entries are 1 offers 1 otherwise 0 || (binary operator): logical OR, if one associated with the final stack entries is 1 result 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 even one 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 to your right
D: duplicates final stack entry 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 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? ) max (binary operator): provides greater of final two stack entries min (binary operator): provides the smaller of last two stack entries
Other people: if if final stack entry is 1, the rule within the brackets is performed (remember that there’s absolutely no SPACE 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 ( just simply simply take also care to where SPACEs are allowed and where perhaps not) quit simply 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 forced along the stack, 7 states exactly just how entries that are much in line with the consequence of (A: Flaps handle index, quantity) ”case” extracts one of several seven figures. If (A: Flaps handle index, quantity) is 0 – 0, 1-1, 2-5. 6-30.