Re: Building a string from n chars
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.
Re: Building a string from n chars
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!? :=)
Re: Building a string from n chars
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); .
Re: Building a string from n chars
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.
Re: Building a string from n chars
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.
Re: Building a string from n chars
On Thursday, 4 September 2014 at 20:57:43 UTC, monarch_dodra wrote: 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 re-read the doc and implementation: replicate replicates a *range*. It is a bit optimized to detect the case where the range is a single element, but it still has to do the check, and even then (implementation detail), it is less efficient. I might create a pull to tweak that.
Re: Building a string from n chars
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);
Re: Building a string from n chars
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.
Re: Building a string from n chars
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 .. $] []= '*';
This is also relatively simple. The ownside is doing double
assignement, as length will initialize new elements to
char.init. Probably not noticeable.
There *might* be more efficient ways to do it, but I'd doubt it
qualifies as simpler. Or if it's really observeable. Are these
good enough for you?
Re: Building a string from n chars
On Wednesday, 3 September 2014 at 20:46:40 UTC, monarch_dodra
wrote:
On Wednesday, 3 September 2014 at 19:43:26 UTC, Nordlöw wrote:
Is there a simpler way to way to
s ~= repeat('*', n).array.to!string;
if s has to be of type string?
s ~= repeat('*', n).array();
Should be enough. Why the to!string?
There's 1 useless allocation, but I think that's OK for code
this trivial?
Else, you can do:
s.length+=n;
s[$-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[].
