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I think PHPs big strength is the other point mentioned in the post: It enabled lots of people who are not really programmers to do some programming. Some of the

by erikb 12y ago

I think PHPs big strength is the other point mentioned in the post: It enabled lots of people who are not really programmers to do some programming. Some of them developed into serious developers, some of them didn't. Being able to produce something slightly meaningful even without any skill is a big accomplishment, but it also means loads of erros, over estimated skill levels, security holes, unreadable code etc. The reason this doesn't happen to Haskel is that you become a computer scientist first and then learn that language. The reason is not a single bad book, where probably most programming languages have a few of.

rfergie 12y ago
The reputation rather than the actual experience of Haskell is what puts people off.
- tormeh 12y ago
  I have never bounced off any language other than Haskell. Lisps are easy to learn by comparison.
  - Ma8ee 12y ago
    Ok, now I really have to learn Haskell.
- erikb 12y ago
  What have you heard? All I've heard makes me sad I never got around to learning me some Haskell.
  - rfergie 12y ago
    Haskell was the first language I learned. There is tremendous depth to it which I am still exploring but to start and do the beginner kind of things I was wanting to do it is not hard. Maybe not as easy as Python, but there is a huge gap between "not quite as easy as" and the "you need a CS degree and a Phd in Category Theory to even get started" rhetoric.
- CJefferson 12y ago
  Here was about my first experience of Haskell. I don't mind admitting I bounced fairly hard at first. I've since done a bit more, but find it too limiting rather than freeing. Prelude> foldr + [1,2,3] <interactive>:14:9: Couldn't match expected type ‘(a -> b -> b) -> b -> [a] -> b’ with actual type ‘[t0]’ Relevant bindings include it :: (a -> b -> b) -> b -> [a] -> b (bound at <interactive>:14:1) In the second argument of ‘(+)’, namely ‘[1, 2, 3]’ In the expression: foldr + [1, 2, 3] In an equation for ‘it’: it = foldr + [1, 2, 3]
  - Viktoras25 12y ago
    Finally I found out the correct version: foldr (+) 0 [1,2,3] 1. + contains only special characters and must be used either as infix function (1 + 2) or be enclosed in brackets 2. You need an initial value for the accumulator. You are reducing a list to a single value, you need to start somewhere Please, gurus of Haskell, correct me if I'm wrong, I've just recently started to learn the language

by erikb 12y ago

rfergie 12y ago

The reputation rather than the actual experience of Haskell is what puts people off.

tormeh 12y ago
I have never bounced off any language other than Haskell. Lisps are easy to learn by comparison.
- Ma8ee 12y ago
  Ok, now I really have to learn Haskell.
erikb 12y ago
What have you heard? All I've heard makes me sad I never got around to learning me some Haskell.
- rfergie 12y ago
  Haskell was the first language I learned. There is tremendous depth to it which I am still exploring but to start and do the beginner kind of things I was wanting to do it is not hard. Maybe not as easy as Python, but there is a huge gap between "not quite as easy as" and the "you need a CS degree and a Phd in Category Theory to even get started" rhetoric.
CJefferson 12y ago
Here was about my first experience of Haskell. I don't mind admitting I bounced fairly hard at first. I've since done a bit more, but find it too limiting rather than freeing. Prelude> foldr + [1,2,3] <interactive>:14:9: Couldn't match expected type ‘(a -> b -> b) -> b -> [a] -> b’ with actual type ‘[t0]’ Relevant bindings include it :: (a -> b -> b) -> b -> [a] -> b (bound at <interactive>:14:1) In the second argument of ‘(+)’, namely ‘[1, 2, 3]’ In the expression: foldr + [1, 2, 3] In an equation for ‘it’: it = foldr + [1, 2, 3]
- Viktoras25 12y ago
  Finally I found out the correct version: foldr (+) 0 [1,2,3] 1. + contains only special characters and must be used either as infix function (1 + 2) or be enclosed in brackets 2. You need an initial value for the accumulator. You are reducing a list to a single value, you need to start somewhere Please, gurus of Haskell, correct me if I'm wrong, I've just recently started to learn the language