4 ms·

Yours doesn't hash unique words though

• Veedrac 11y ago
  The original implementation's bucketing was completely irrelevant to the goal - it still ended up counting the number of words, duplicates included.
  • lqdc13 11y ago
    They're not counting duplicates though... Both perl and C++ one use a hash table.
    • Veedrac 11y ago
      They use a <word → word frequency> mapping, and then sum the values. This gives the total number of words. The hash table is totally pointless.
      • lqdc13 11y ago
        You're totally right.
      • draegtun 11y ago
        The author did have a point he wanted to put across with the hash table - "Obviously, a frequency distribution was not necessary for just the counts, but I did want to highlight basic autovivification" ref: http://www.reddit.com/r/cpp/comments/369lcn/can_c_become_your_new_scripting_language/crcklhr http://www.reddit.com/r/cpp/comments/369lcn/can_c_become_you...
        • Veedrac 11y ago
          That doesn't feel like a good reason. The author could easily have come up with a better task (eg. count the number of words with more than one occurrence). This allows you to approach the problem in the optimal way in each language.
      • colomon 11y ago
        Ooo, I can do that in Perl 6 too. #!/usr/bin/env perl6 sub MAIN(*@filename) { for @filename -> $file { my $bag = $file.IO.slurp.words.Bag; say "$file: { [+] $bag.values } words"; CATCH { default { say "Unable to count words in $file"; } } } }