findBack: find a needle in a haystack from the back
Tobias Pankrath via Digitalmars-d-learn
digitalmars-d-learn at puremagic.com
Mon Jun 9 04:56:44 PDT 2014
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?
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