Re: Alternate signs in a range

[Stanislav Blinov]

But it's easy to avoid the multiplication and replace it with 
a conditional neg.

Bye,
bearophile

How would you do this ?

map!(a => (a.index & 1) ? a : -a)([1, 2]) ? (index field isn't 
available of course)

import std.range;
import std.algorithm;
import std.stdio;

void main() {

        auto myRange = iota(0,10);

        auto index = sequence!((a,n) => a[0]+n)(1);
	auto result = myRange.zip(index).map!(a => a[1] & 1 ? a[0] : 
-a[0]);
        result.writeln;
}