(From memory & without firing up a compiler to verify) ...
Sets are implemented as bitmaps with one bit allocated to each possible
value of the set type (hence the limitations on the number of possible
elements in the set).
So if you have
type bytesset = [0..7]; // 8 possible values
var a: bytesset ;
begin
if a in [1,4] then
The set literal [1,4] ends up as a byte with the value $0A.
The test "a in [1,4]" is equivalent to "(a AND $0A) <> 0".
The generated code (for a set of this size) is an AND followed by a
conditional jump. Very efficient.
It gets a little nastier when the possible values gets larger as the
size of the set grows (set of AnsiChar is 32 bytes; Set of Widechar
exceeds the permissible size of a set).
The precise code generated by case varies between jump tables and linear
comparisons.
Regards
On 8/06/2016 5:27 p.m., Steve Peacocke wrote:
If I remember correctly, the compiler changes both to array anyway so
you come out with exactly the same compiled code.
Perhaps someone can confirm this or tell me how wrong I am?
Steve Peacocke
+64 220 612-611
On 8/06/2016, at 4:56 PM, Ross Levis <[email protected]
<mailto:[email protected]>> wrote:
I’m wondering which is more efficient to process...
if (a=1) or (a=2) then ...
or
if a in [1,2] then ...
If the answer is the first method, does it make a difference if more
numbers are checked, eg. if a in [1..3,5] then
Cheers.
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