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digitalmars.D.learn - Building a string from n chars

reply =?UTF-8?B?Ik5vcmRsw7Z3Ig==?= <per.nordlow gmail.com> writes:
Is there a simpler way to way to

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

if s has to be of type string?
Sep 03 2014
next sibling parent reply "Meta" <jared771 gmail.com> writes:
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);
Sep 03 2014
next sibling parent "Meta" <jared771 gmail.com> writes:
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.
Sep 03 2014
prev sibling parent reply =?UTF-8?B?Ik5vcmRsw7Z3Ig==?= <per.nordlow gmail.com> writes:
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!?

:=)
Sep 04 2014
parent reply =?UTF-8?B?Ik5vcmRsw7Z3Ig==?= <per.nordlow gmail.com> writes:
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);

.
Sep 04 2014
parent reply =?UTF-8?B?Ik5vcmRsw7Z3Ig==?= <per.nordlow gmail.com> writes:
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 


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.
Sep 04 2014
parent reply "monarch_dodra" <monarchdodra gmail.com> writes:
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 


 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.
Sep 04 2014
parent "monarch_dodra" <monarchdodra gmail.com> writes:
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 


 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.
Sep 04 2014
prev sibling parent reply "monarch_dodra" <monarchdodra gmail.com> writes:
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?
Sep 03 2014
next sibling parent "Cassio Butrico" <cassio_butrico ig.com.br> writes:
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[].
Sep 03 2014
prev sibling parent =?UTF-8?B?Ik5vcmRsw7Z3Ig==?= <per.nordlow gmail.com> writes:
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.
Sep 04 2014