Re: scalar in list context

Maybe you mean the context of the assignment?

@array = ( "one", "two", "three" );
($len) = @array;
$len1 = @array;

print "1 - ", $len, "\n";
print "2 - ", $len1, "\n";
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Where the first will print "one" and the second will print "3", since (IIRC) assignment forces the context of the LHS onto the RHS.

Update: See ikegami's post below for the correct explanation of what I was trying to say.