On Thu, May 17, 2012 at 5:05 PM, Haiyan Lin <linhy0120@gmail.com> wrote:
Hi, perlers,--
Supposing I have a hash reference in hand, the effect of the following two statements
$hash1_ref = &subroutine1( $hash0_ref, 'parameter1' );
$hash2_ref = &subroutine1( $hash1_ref, 'parameter2' );
will save elements in $hash0_ref, stafisfying both parameter1 and parameter2, into $hash2_ref.
I try to do this by following one statement, but I failed.
$hash_ref2 = &subroutine1( &subroutine1( $hash0_ref, 'parameter1'), 'paratemter2') ;
any help? THANKS in advance
==================================================
下面的语句可以得到$hash0_ref中同时满足parameter1 和parameter2的结果,并放到$hash2_ref中。
$hash1_ref = &subroutine1( $hash0_ref, 'parameter1' );
$hash2_ref = &subroutine1( $hash1_ref, 'parameter2' );
现在我换一种写法,理论上,可以得到相同的结果,但报错了。请高高手指导。谢谢
$hash_ref2 = &subroutine1( &subroutine1( $hash0_ref, 'parameter1'), 'paratemter2') ;
Haiyan Lin
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--
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