Is there a simpler way to way to

s ~= repeat('*', n).array.to!string;

if s has to be of type string?

On Wednesday, 3 September 2014 at 19:43:26 UTC, Nordlöw wrote:
> Is there a simpler way to way to
>
> s ~= repeat('*', n).array.to!string;
>
> if s has to be of type string?

Does this work?

s ~= "*".replicate(n);

On Wednesday, 3 September 2014 at 20:20:09 UTC, Meta wrote:
> On Wednesday, 3 September 2014 at 19:43:26 UTC, Nordlöw wrote:
>> Is there a simpler way to way to
>>
>> s ~= repeat('*', n).array.to!string;
>>
>> if s has to be of type string?
>
> Does this work?
>
> s ~= "*".replicate(n);

Sorry, I should qualify that replicate is from std.array. You can also just do either `s ~= repeat('*', n).array` OR `s ~= repeat('*', n).to!string`. Either should work.

On Wednesday, 3 September 2014 at 19:43:26 UTC, Nordlöw wrote:
> Is there a simpler way to way to
>
> s ~= repeat('*', n).array.to!string;
>
> if s has to be of type string?

s ~= repeat('*', n).array();

Should be enough. Why the "to!string"?

There's 1 useless allocation, but I think that's OK for code this trivial?

Else, you can do:
s.length+=n;
s[$-n .. $] []= '*';

This is also relatively simple. The ownside is doing double assignement, as "length" will initialize new elements to "char.init". Probably not noticeable.

There *might* be more "efficient" ways to do it, but I'd doubt it qualifies as "simpler". Or if it's really observeable. Are these good enough for you?

On Wednesday, 3 September 2014 at 20:46:40 UTC, monarch_dodra wrote:
> On Wednesday, 3 September 2014 at 19:43:26 UTC, Nordlöw wrote:
>> Is there a simpler way to way to
>>
>> s ~= repeat('*', n).array.to!string;
>>
>> if s has to be of type string?
>
> s ~= repeat('*', n).array();
>
> Should be enough. Why the "to!string"?
>
> There's 1 useless allocation, but I think that's OK for code this trivial?
>
> Else, you can do:
> s.length+=n;
> s[$-n .. $] []= '*';
>
> This is also relatively simple. The ownside is doing double assignement, as "length" will initialize new elements to "char.init". Probably not noticeable.
>
> There *might* be more "efficient" ways to do it, but I'd doubt it qualifies as "simpler". Or if it's really observeable. Are these good enough for you?

hello Nordlöw a time I did something like that, see.
string repet(string a, int i)
	{
		int b;
		string c;
		for(b=0;b<=i;b++)
		{c ~= a;}
		return c;
	}

can be adapted to char[].

On Wednesday, 3 September 2014 at 20:46:40 UTC, monarch_dodra wrote:
>> Is there a simpler way to way to
>>
>> s ~= repeat('*', n).array.to!string;
>>
>> if s has to be of type string?
>
> s ~= repeat('*', n).array();
>
> Should be enough. Why the "to!string"?

You're correct. The

    .to!string

was unnecessary.

On Wednesday, 3 September 2014 at 20:20:09 UTC, Meta wrote:
> Does this work?
>
> s ~= "*".replicate(n);

Yes, thanks.

So what's best?

    type ~= '*'.repeat(pointerCount).array;

or

    type ~= "*".replicate(pointerCount);

?

Further, -vgc says only ~= will allocate:

t_repeat_replicate.d(12,19): vgc: operator ~= may cause GC allocation
t_repeat_replicate.d(13,19): vgc: operator ~= may cause GC allocation

Is DMD/Phobos already that clever!?

:=)

On Thursday, 4 September 2014 at 19:22:42 UTC, Nordlöw wrote:
> Is DMD/Phobos already that clever!?

Further -vgc has nothing to say about

    string t1; t1 ~= '*'.repeat(n).array;
    string t2; t2 ~= "*".replicate(n);

.

On Thursday, 4 September 2014 at 19:24:00 UTC, Nordlöw wrote:
>     string t1; t1 ~= '*'.repeat(n).array;
>     string t2; t2 ~= "*".replicate(n);

After having read http://dlang.org/phobos/std_array.html#.replicate

I came to the conclusion that the lazy std.range:repeat is preferred.

I'm still a bit confused about the fact that -vgc gives no warnings about GC-allocations, though.

On Thursday, 4 September 2014 at 20:38:39 UTC, Nordlöw wrote:
> On Thursday, 4 September 2014 at 19:24:00 UTC, Nordlöw wrote:
>>    string t1; t1 ~= '*'.repeat(n).array;
>>    string t2; t2 ~= "*".replicate(n);
>
> After having read http://dlang.org/phobos/std_array.html#.replicate
>
> I came to the conclusion that the lazy std.range:repeat is preferred.

If lazy is good enough for you yes. AFAIK, replicate is *very* close in terms of implementation to what a.repeat(n).array() would do anyways. Heck, I'd be surprised if it did it differently, since (again, AFAIK) repeat.array() is "optimal" anyways.

> I'm still a bit confused about the fact that -vgc gives no warnings about GC-allocations, though.

Strange indeed. Both solutions allocate a slice, and then append that slice. The "s[]='*'" Solution I gave you will not create a temporary allocation.