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Old 23rd September 2010
cq04cw cq04cw is offline
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Join Date: Sep 2010
Posts: 5

Thanks, rocket357.

The first form b=$( eval echo \$$_var ) which i understood is parsed like this:
pass1: eval echo $aa ## \$ is treated as dollar sign $, $_var is aa
pass2: eval it , and assign result (echo $aa 's output) to variable b
is it right to think of it ?

however, the second form b=`eval echo \\$$_var` which i can't read it.
pass1: eval echo \$aa ## \\ -> \ (as manpage explained), $_var -> aa
pass2: eval it ? echo \$aa -> $aa , $aa is output of echo , b=$aa, but the result is b=AAAA ($aa's val).

how to understand the second form ?
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