commonLength

[Ali Çehreli via Digitalmars-d-learn]

On 01/15/2015 11:24 AM, "Nordlöw" wrote:
> On Thursday, 15 January 2015 at 13:13:57 UTC, Nordlöw wrote:
>> I just discovered that zip has StoppingPolicy so why does
>>
>> auto commonPrefixLength(R...)(R ranges) if (ranges.length == 2)
>> {
>>     import std.range: zip;
>>     return zip!((a, b) => a[0] != b[1])(ranges);
>> }
>
> I did a silly mistake. The correct version is
>
> auto commonPrefixLength(S, T)(S a, T b)
> {
>      import std.range: zip, StoppingPolicy;
>      import std.algorithm: countUntil, count;
>      const hit = zip(a, b).countUntil!(ab => ab[0] != ab[1]);
>      return hit == -1 ? zip(a, b).count : hit;
> }
>
> This however needs to process zip(a, b) how do I avoid the extra count?
>
> If countUntil returned zip(a, b).count upon failure I would have been
> done...

The predicate version of count() seems to work for your case. If I 
misunderstood, please explain with an added failing unit test:

auto commonPrefixLength(S, T)(S a, T b)
{
     import std.range: zip;
     import std.algorithm: count;
     return zip(a, b).count!(ab => ab[0] == ab[1]);
}

unittest
{
     assert(commonPrefixLength("açde", "") == 0);
     assert(commonPrefixLength("açde", "xyz") == 0);
     assert(commonPrefixLength("açd", "açde") == 3);
     assert(commonPrefixLength("açdef", "açdef") == 5);
}

void main()
{}

Ali