In crystal Why do I need to do matches.try &.[0]












1















Here's some code:



matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address) # line 1
puts matches[0]? # line 2
puts matches.try &.[0] # line 3


matches datatype is (Regex::MatchData | Nil) according to typeof. Yet, line 2, referring to matches[0] fails with a compilation error. And I don't understand line 3 at all!



Could someone clarify?










share|improve this question





























    1















    Here's some code:



    matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address) # line 1
    puts matches[0]? # line 2
    puts matches.try &.[0] # line 3


    matches datatype is (Regex::MatchData | Nil) according to typeof. Yet, line 2, referring to matches[0] fails with a compilation error. And I don't understand line 3 at all!



    Could someone clarify?










    share|improve this question



























      1












      1








      1








      Here's some code:



      matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address) # line 1
      puts matches[0]? # line 2
      puts matches.try &.[0] # line 3


      matches datatype is (Regex::MatchData | Nil) according to typeof. Yet, line 2, referring to matches[0] fails with a compilation error. And I don't understand line 3 at all!



      Could someone clarify?










      share|improve this question
















      Here's some code:



      matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address) # line 1
      puts matches[0]? # line 2
      puts matches.try &.[0] # line 3


      matches datatype is (Regex::MatchData | Nil) according to typeof. Yet, line 2, referring to matches[0] fails with a compilation error. And I don't understand line 3 at all!



      Could someone clarify?







      crystal-lang






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 20 '18 at 4:49









      Cœur

      18.2k9108148




      18.2k9108148










      asked Nov 20 '18 at 1:37









      pitosalaspitosalas

      3,24063365




      3,24063365
























          1 Answer
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          4














          Line 2: As you say, the type of matches is (Regex::MatchData | Nil). In case it happens to be nil, it wouldn't have the #? method, which makes the type checker angry. You are supposed to check whether match succeeded first:



          matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address)
          if matches
          puts matches[0]?
          end


          Inside if, the type of matches is just Regex::MatchData (as we eliminated the Nil possibility), and the type checker can rest peacefully.



          If you are sure your string will match, you can pacify the type checker with not_nil!, but that opens up a possibility of a runtime error, if your confidence in your data's conformance was unfounded:



          puts matches.not_nil![0]?


          Line 3: #try will perform the block, except when the caller is nil, when it returns nil. No guards are needed because #try is explicitly defined on Nil (as well as on Object).



          It uses the shortcut syntax for blocks, where &.[0] is kind of equivalent to { |x| x[0] }.






          share|improve this answer

























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            1 Answer
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            4














            Line 2: As you say, the type of matches is (Regex::MatchData | Nil). In case it happens to be nil, it wouldn't have the #? method, which makes the type checker angry. You are supposed to check whether match succeeded first:



            matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address)
            if matches
            puts matches[0]?
            end


            Inside if, the type of matches is just Regex::MatchData (as we eliminated the Nil possibility), and the type checker can rest peacefully.



            If you are sure your string will match, you can pacify the type checker with not_nil!, but that opens up a possibility of a runtime error, if your confidence in your data's conformance was unfounded:



            puts matches.not_nil![0]?


            Line 3: #try will perform the block, except when the caller is nil, when it returns nil. No guards are needed because #try is explicitly defined on Nil (as well as on Object).



            It uses the shortcut syntax for blocks, where &.[0] is kind of equivalent to { |x| x[0] }.






            share|improve this answer






























              4














              Line 2: As you say, the type of matches is (Regex::MatchData | Nil). In case it happens to be nil, it wouldn't have the #? method, which makes the type checker angry. You are supposed to check whether match succeeded first:



              matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address)
              if matches
              puts matches[0]?
              end


              Inside if, the type of matches is just Regex::MatchData (as we eliminated the Nil possibility), and the type checker can rest peacefully.



              If you are sure your string will match, you can pacify the type checker with not_nil!, but that opens up a possibility of a runtime error, if your confidence in your data's conformance was unfounded:



              puts matches.not_nil![0]?


              Line 3: #try will perform the block, except when the caller is nil, when it returns nil. No guards are needed because #try is explicitly defined on Nil (as well as on Object).



              It uses the shortcut syntax for blocks, where &.[0] is kind of equivalent to { |x| x[0] }.






              share|improve this answer




























                4












                4








                4







                Line 2: As you say, the type of matches is (Regex::MatchData | Nil). In case it happens to be nil, it wouldn't have the #? method, which makes the type checker angry. You are supposed to check whether match succeeded first:



                matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address)
                if matches
                puts matches[0]?
                end


                Inside if, the type of matches is just Regex::MatchData (as we eliminated the Nil possibility), and the type checker can rest peacefully.



                If you are sure your string will match, you can pacify the type checker with not_nil!, but that opens up a possibility of a runtime error, if your confidence in your data's conformance was unfounded:



                puts matches.not_nil![0]?


                Line 3: #try will perform the block, except when the caller is nil, when it returns nil. No guards are needed because #try is explicitly defined on Nil (as well as on Object).



                It uses the shortcut syntax for blocks, where &.[0] is kind of equivalent to { |x| x[0] }.






                share|improve this answer















                Line 2: As you say, the type of matches is (Regex::MatchData | Nil). In case it happens to be nil, it wouldn't have the #? method, which makes the type checker angry. You are supposed to check whether match succeeded first:



                matches =  //([a-z]+)/(d+)/state/([a-z]+)/.match(address)
                if matches
                puts matches[0]?
                end


                Inside if, the type of matches is just Regex::MatchData (as we eliminated the Nil possibility), and the type checker can rest peacefully.



                If you are sure your string will match, you can pacify the type checker with not_nil!, but that opens up a possibility of a runtime error, if your confidence in your data's conformance was unfounded:



                puts matches.not_nil![0]?


                Line 3: #try will perform the block, except when the caller is nil, when it returns nil. No guards are needed because #try is explicitly defined on Nil (as well as on Object).



                It uses the shortcut syntax for blocks, where &.[0] is kind of equivalent to { |x| x[0] }.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 20 '18 at 9:06

























                answered Nov 20 '18 at 8:47









                AmadanAmadan

                130k13143195




                130k13143195
































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