But in Perl 5 `.` is string concatenation operator, not `+`
On 2017-01-13 09:33:10 GMT, Todd Chester wrote: >> >> On Fri, Jan 13, 2017 at 10:01 AM, Todd Chester <toddandma...@zoho.com >> <mailto:toddandma...@zoho.com>> wrote: >> >> Hi All, >> >> I am trying to understand how to read variables from the >> shell's environment. I am reading this: >> >> https://docs.perl6.org/language/variables#Dynamic_variables >> <https://docs.perl6.org/language/variables#Dynamic_variables> >> >> <code> >> #!/usr/bin/perl6 >> # print "Display = " + %*ENV{'DISPALY'} + "\n"; >> print "Perl Version = " + $*PERL + "\n"; >> </code> >> >> >> $ ./env.pl6 >> Cannot resolve caller Numeric(Perl: ); none of these signatures match: >> (Mu:U \v: *%_) >> in block <unit> at ./env.pl6 line 3 >> >> >> What am I missing? I'd like to get the one I commented >> out on line 2 working too. That gives me the following error: >> >> Use of uninitialized value of type Any in numeric context >> in block <unit> at ./env.pl6 line 2 >> Cannot convert string to number: base-10 number must begin with >> valid digits or '.' in '⏏Display = ' (indicated by ⏏) >> in block <unit> at ./env.pl6 line 2 >> >> >> Many thanks, >> -T >> >> >> >> >> -- >> Fernando Santagata > > On 01/13/2017 01:20 AM, Fernando Santagata wrote: >> Try: >> >> print "Perl Version = " ~ $*PERL ~ "\n"; >> >> or better: >> >> say "Perl Version = $*PERL"; >> >> OTH > > > Hi Fernando, > > That fixed it. I was using Perl5's string concatenation. Will "~" > always replace "+" for this, or only with a dynamic variable? > > On line two, I misspelled "DISPLAY". > > <code> > #!/usr/bin/perl6 > print "Display = " ~ %*ENV{'DISPLAY'} ~ "\n"; > print "Perl Version = " ~ $*PERL ~ "\n"; > </code> > > ./env.pl6 > Display = :0.0 > Perl Version = Perl 6 > > > Thank you! > -T
