Re: How to sort byCodeUnit.permutations.filter(...)

[Adam D. Ruppe via Digitalmars-d-learn]

On Monday, 11 June 2018 at 04:39:54 UTC, Uknown wrote:
Why are the strings getting modified?

I'm guessing it reuses a buffer as it iterates.

"123".byCodeUnit.permutations.writeln;//[123, 213, 312, 132, 
231, 321]

so here, it switches them, prints, switches, prints, switches, 
prints,e tc

"123".byCodeUnit.permutations.array.writeln;//[123, 123, 123, 
123, 123, 123]

but here it adds the same pointer to the array over and over so 
it all points to the same buffer.

I'm guessing. I don't actually know.


But hmmm... maybe throwing in a .dup would help. Really though, I 
personally would forget all the range pipeline stuff and write a 
simple loop.


        int largest = 0;
        foreach(part; "1234567".byCodeUnit.permutations) {
             auto asNumber = to!int(part);
             if(primes.canFind(asNumber)) {
                if(asNumber > largest)
                   largest = asNumber;
             }
        }

        writeln(largest);


But, writing the simple loop did give me a solution to the 
pipeline too: map it to to!int before doing the rest, so you work 
with ints instead of with strings and char buffers:


        "1234567".byCodeUnit
                .permutations
                .map!(to!int)
                .filter!(a => primes.canFind(a))
                .maxElement


that should do it too.