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bash$ a=5
bash$ b=a
bash$ eval echo "$"$b
5

P.S. Думаю можно как-то и попроще :-)

spirit ★★★★★
()
Ответ на: комментарий от LowLevel

>кульно. а как это расшифровывается?
man bash :)

       ${parameter}
              The value of parameter is substituted.  The braces are required when parameter is a positional  parameter  with  more  than  one
              digit, or when parameter is followed by a character which is not to be interpreted as part of its name.

       If the first character of parameter is an exclamation point, a level of variable indirection is introduced.  Bash uses the value of the
       variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is  used  in  the
       rest  of the substitution, rather than the value of parameter itself.  This is known as indirect expansion.  The exceptions to this are
       the expansions of ${!prefix*} and ${!name[@]} described below.  The exclamation point must immediately follow the left brace  in  order
       to introduce indirection.

rip_someday
()
Ответ на: комментарий от rip_someday

понятно. хак.

нет бы по человечески - ${$b} =)

LowLevel
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