how to avoid array pop effect ?

@aQuosFeb 08.2007

How can i avoid the array’s pop effect that executes pop on array1 too ?

[CODE]<script type=”text/javascript”> var array1 = new Array() ; var array2 = new Array() ; array1[0] = ‘foo’ ; array1[1] = ‘miao’ ; array1[2] = ‘grrr’ ; array2 = array1 ; array2.pop() ; alert ( array1 ) ; alert ( array2 ) ; </script>[/CODE]

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JavaScript

@konithomimoFeb 08.2007 — #Well, you will always get that problem if you set them equal. To fix it you can populate array2 with a simple for loop:

for(var i=0;i&lt;array1.length;i++)
 array2[i] = array1[i];

Which makes the entire code:

&lt;script type="text/javascript"&gt;
 var array1 = new Array() ;
 var array2 = new Array() ;
 array1[0] = 'foo' ;
 array1[1] = 'miao' ;
 array1[2] = 'grrr' ;
 for(var i=0;i&lt;array1.length;i++)
 array2[i] = array1[i];
 
 array2.pop() ;
 alert ( array1 ) ;
 alert ( array2 ) ;
 &lt;/script&gt;

@gil_davisFeb 08.2007 — #This is rather silly, but effective:

<i>
 </i>var array1 = new Array() ;
 var array2 = new Array() ;
 array1[0] = 'foo' ;
 array1[1] = 'miao' ;
 array1[2] = 'grrr' ;
 [color=red]//[/color]array2 = array1 ;
 [color=red]var t = array1.join("~");
 array2 = t.split("~");[/color]
 array2.pop() ;
 alert ( array1 ) ;
 alert ( array2 ) ;

@aQuosauthorFeb 08.2007 — #nice! thanks!!

how to avoid array pop effect ?

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