why do I have to provide context to split result in a separate statement?

Hello, why does this bomb

print @{\(@arr = split / /, '')}
[download]

while this does not

@arr = split / /, '';
print @{\@arr}
[download]

Thank you, Mark

Comment on why do I have to provide context to split result in a separate statement? Select or Download Code
Re: why do I have to provide context to split result in a separate statement? by jwkrahn (Prior) on Sep 21, 2011 at 01:40 UTC
`\(@arr = split / /, '')` returns a list of references. For example `\( $a, $b, $c )` is actually `( \$a, \$b, \$c )` and you can't dereference that as an array. What you could do is copy the list to an anonymous array: `print @{[@arr = split / /, '']}` [download]	[reply] [d/l] [select]
Re^2: why do I have to provide context to split result in a separate statement? by ikegami (Pope) on Sep 21, 2011 at 01:43 UTC
Don't need to copy it into two arrays. `print @{[ split / /, '' ]}` [download]	[reply] [d/l]
Re^3: why do I have to provide context to split result in a separate statement? by jwkrahn (Prior) on Sep 21, 2011 at 01:59 UTC
Tell that to Mark_Galeck `:-)`	[reply] [d/l]
Re^4: why do I have to provide context to split result in a separate statement? by ikegami (Pope) on Sep 21, 2011 at 02:11 UTC
Re^3: why do I have to provide context to split result in a separate statement? by Mark_Galeck (Novice) on Sep 21, 2011 at 02:04 UTC
Thank you very much!	[reply]


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