On Saturday, 23 April 2016 at 20:01:00 UTC, ag0aep6g wrote:
On 23.04.2016 21:49, xtreak wrote:
I am a D newbie from Python and I am trying to grok alias. Is
alias like
Python does as below
L = []
myextend = L.extend
L.myextend
My Python isn't too great, but I think this is more similar to
function pointers or delegates in D.
Renaming imported function
from itertools import permutations as p
p([1, 2], 2)
Yes, that's similar. A renamed import creates an alias. For
example, `import std.algorithm: p = permutations;` creates an
alias `p` for std.algorithm.permutations.
Is D aliasing the same as above? How does aliasing types help
like below
alias intList = LinkedList!int
Is the above like a partially applied template as in
LinkedList!int([1,
2, 3]) and hence can I use it like intList([1, 2, 3])?
No, the template isn't partially applied, it's fully
instantiated (and results in a type). The alias declaration
just makes `intList` an alternative name for `LinkedList!int`.
import std.array;
import std.range;
import std.algorithm;
import std.stdio;
T test(alias f, T)(T num) {
return f(num);
}
T test1(T, V)(T num, V f){
return f(num);
}
void main() {
writeln("hello world");
writeln(10000.iota
.map!(a => a * a)
.take(5));
writeln(test!(z => z * z)(10));
writeln(test1(10, ((z) => z *z)));
writeln(test1(10, function int(int z) { return z * z; }));
}
What is the difference between passing as alias and then passing
it as lambda. Does it involve any cost. Also the second form of
short notation throws an error that it returns void. Kindly help
me on this as now alias is not pointing to a named symbol so is
there any cost and why alias is preferred
