rewrite *p++=*q++ in haskell?
it's one of C idioms. probably, you don't have enough C experience to understand it :)
Maybe, but how can *you* understand it, when the standard is vague about it? It could be A: *p=*q; p+=1; q+=1; B: *p=*q; q+=1; p+=1; C: tp=p; tq=q; p+=1; q+=1; *tp=*tq; ...and so on. Which is the "right" version?
result is that currently C code rewritten in Haskell becomes much larger and less readable.
Larger should not be that issue and readability depends on the reader as your C example shows. Some Haskellers would very quickly recognize some common idioms, where others need some help...
/BR -- -- Mirko Rahn -- Tel +49-721 608 7504 -- --- http://liinwww.ira.uka.de/~rahn/ --- _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
