On Tue, 2010-10-05 at 15:46 -0400, Steve Staples wrote: > On Tue, 2010-10-05 at 20:35 +0100, Ashley Sheridan wrote: > > On Tue, 2010-10-05 at 15:28 -0400, chris h wrote: > > > > > Benchmark and find out! :) > > > > > > What are you using this for? Unless you are doing something crazy it > > > probably doesn't matter, and you should pick whichever you feel looks nicer > > > / is easier to code in / etc. > > > > > > Chris H. > > > > > > On Tue, Oct 5, 2010 at 3:23 PM, saeed ahmed <saeed.sas@xxxxxxxxx> wrote: > > > > > > > $a = 'hey'; > > > > $b = 'done'; > > > > > > > > $c = $a.$b; > > > > $c = "$a$b"; > > > > > > > > which one is faster for echo $c. > > > > > > > > > > As far as I'm aware, the first of the two will be faster, but only just. > > As Saeed mentioned, the difference will be negligible, and unless you > > plan to run a line like that in a loop or something hundreds of > > thousands of times, you probably won't notice any difference. > > Thanks, > > Ash > > http://www.ashleysheridan.co.uk > > > > > > > to be proper, shouldn't it technically be > $c = "{$a}{$b}"; > > ?? > > Steve. > > It doesn't have to use the braces. The braces only tell PHP exactly where to stop parsing the current variable name. The following examples wouldn't work without them: $var = 'hello '; $arr = array('msg 1'=>'hello','msg 2'=>'world'); echo "{$var}world"; echo "{$arr['msg 1']}{$arr['msg 2']}"; Without the braces, in the first example PHP would look for a variable called $varworld, and in the second it would be looking for a simple scaler called $arr, not the array value you wanted. Thanks, Ash http://www.ashleysheridan.co.uk
