• bearophile (21/21) Apr 11 2014 If I have a range like this, how do you find its last item using
• Steven Schveighoffer (5/10) Apr 11 2014 Interesting problem. Given that it is a forward range, a zip between it ...
• bearophile (11/14) Apr 12 2014 And it's not an academic one, I have already faced it two or
• monarch_dodra (25/34) Apr 12 2014 For starters, you can't in the sense that writeln takes by value,

"bearophile" <bearophileHUGS lycos.com> writes:

If I have a range like this, how do you find its last item using 
Phobos?

A simple way is to use seq.array.back, but this wastes memory. 
Another simple way is to use a foreach loop keeping the last 
seen. But do you know if there is a function to do in Phobos? And 
if such function is not present (walkBack? backWalk?) is it a 
good idea to add it to Phobos?


void main() {
     import std.stdio, std.algorithm, std.range;

     int m = 75;
     auto seq = recurrence!q{ a[n - 1] + a[n - 2] }(1, 1)
                .until!(x => x > m)(OpenRight.no);

     seq.array.back.writeln; // 89
     //seq.back.writeln;

     auto last = -1;
     foreach (immutable x; seq)
         last = x;
     last.writeln;
}


Bye,
bearophile

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

On Fri, 11 Apr 2014 21:05:26 -0400, bearophile <bearophileHUGS lycos.com>  
wrote:

 If I have a range like this, how do you find its last item using Phobos?

 A simple way is to use seq.array.back, but this wastes memory. Another  
 simple way is to use a foreach loop keeping the last seen. But do you  
 know if there is a function to do in Phobos? And if such function is not  
 present (walkBack? backWalk?) is it a good idea to add it to Phobos?

Interesting problem. Given that it is a forward range, a zip between it  
and a saved copy that is advanced by one may work.

-Steve

"bearophile" <bearophileHUGS lycos.com> writes:

Steven Schveighoffer:

 Interesting problem.

And it's not an academic one, I have already faced it two or 
three times :-)


 Given that it is a forward range, a zip between it and a saved 
 copy that is advanced by one may work.

This compiles and gives the expected output, but how do you 
extract the last item with this?

seq.zip(seq.dropOne).writeln;

I have written I've added an ER to Phobos with some preliminary 
code:
https://issues.dlang.org/show_bug.cgi?id=12564

Bye,
bearophile

"monarch_dodra" <monarchdodra gmail.com> writes:

On Saturday, 12 April 2014 at 10:27:00 UTC, bearophile wrote:
 Steven Schveighoffer:

 Interesting problem.

 And it's not an academic one, I have already faced it two or 
 three times :-)


 Given that it is a forward range, a zip between it and a saved 
 copy that is advanced by one may work.

 This compiles and gives the expected output, but how do you 
 extract the last item with this?

 seq.zip(seq.dropOne).writeln;

For starters, you can't in the sense that writeln takes by value, 
and your zip has value semantics. So in fact, it is still 
unchanged. BTW: That UFCS "seq.zip(someOtherRange)": Ew.

So you'd need to start with a:
auto z = zip(seq, seq.dropOne);
while (!z.empty)
     z.popFront();

That's step one. Step two would be actually extracting the 
remainaing range. In theory, you can't. In practice, you can hack 
around the implementation by:
zip.ranges[1].front;

Turns out "ranges" is not private. I wouldn't use this.

The second, is modifying the stopping policy on the fly:
zip.stoppingPolicy = StoppingPolicy.longest;
auto a = zip.front[1];

I think both are *very* hackish.
I'd do this:

auto bck = seq.save; seq.popFront();
while (!seq.empty)
{
     bck.popFront();
     seq.popFront();
}
auto last = bck.front;