Is there a more idiomatic/elegant way to achieve the following, while
remaining as efficient as possible?
Same question in the simpler case n==0?

using retro seems inefficient because of all the decodings

// returns the largest suffix of a that contains no more than n times c
string findBack(string a, char c, size_t n=0){
  auto b=cast(immutable(ubyte)[])a;
  auto val=cast(ubyte)c;
  size_t counter=0;
  for(ptrdiff_t i=cast(ptrdiff_t)b.length - 1; i>=0; i--){
    if(b[i]==c){
      if(counter>=n)
        return cast(string)a[i+1..$];
      counter++;
    }
  }
  return a;
}

//unittest{
void test3(){
  auto c='\n';
  string a="abc1\nabc2\nabc3";
  assert(a.findBack(c,0)=="abc3");
  assert(a.findBack(c,1)=="abc2\nabc3");
  assert(a.findBack(c,2)=="abc1\nabc2\nabc3");
  assert(a.findBack(c,3)=="abc1\nabc2\nabc3");
}

On Monday, 9 June 2014 at 07:58:25 UTC, Timothee Cour via Digitalmars-d-learn wrote:
> Is there a more idiomatic/elegant way to achieve the following, while
> remaining as efficient as possible?
> Same question in the simpler case n==0?
>
> using retro seems inefficient because of all the decodings
>
> // returns the largest suffix of a that contains no more than n times c
> string findBack(string a, char c, size_t n=0){
>   auto b=cast(immutable(ubyte)[])a;
>   auto val=cast(ubyte)c;
>   size_t counter=0;
>   for(ptrdiff_t i=cast(ptrdiff_t)b.length - 1; i>=0; i--){
>     if(b[i]==c){
>       if(counter>=n)
>         return cast(string)a[i+1..$];
>       counter++;
>     }
>   }
>   return a;
> }
>
> //unittest{
> void test3(){
>   auto c='\n';
>   string a="abc1\nabc2\nabc3";
>   assert(a.findBack(c,0)=="abc3");
>   assert(a.findBack(c,1)=="abc2\nabc3");
>   assert(a.findBack(c,2)=="abc1\nabc2\nabc3");
>   assert(a.findBack(c,3)=="abc1\nabc2\nabc3");
> }

If you are going to cast the string to ubyte[] anyway, why not just do this before using retro?

> using retro seems inefficient because of all the decodings

Phobos git master just got support for next-gen string processing:

https://github.com/D-Programming-Language/phobos/pull/2043

I believe

x.byChar.retro

is what you want