Why does any programming language have infix operators? Because they make some things easier to read. Fixity and precedence are not accidental complexity. Th

Why does any programming language have infix operators? Because they make some things easier to read. Fixity and precedence are not accidental complexity. They are the essential complexity of having infix operators. In order to evaluate "3 * 4 ^ 3 ^ 2 / 4 / 2 + 5 * 8" you must define the fixity and precedence of the operators involved. You don't usually have to think about defining these things in most mainstream languages because the language defines the fixity and precedence of those operators for you to give the behavior you expect. You still have to learn it though. Do you know how Python evaluates the above expression? (Change the carets to double stars...HN didn't render double star properly.) In my first programming class long before Haskell was created, the textbook had a C operator precedence chart. The concept is not new to Haskell. Haskell has just given you control over something that has been there the whole time. Haskell lets you make your own infix operators so that you can get the same benefits in other domains than the ones that the language designers decided to provide for you. Therefore it follows that Haskell also needs to allow you to specify the fixity and precedence so that your operators will fit together in the way that is appropriate for your domain. > Let me explain it a bit further: to read someone's code, first I have to see what symbols they have defined and what fixity they have ascribed to each of those. Now I have to keep all that additional unnecessary cognitive load in my brain to just be able to get a hang of their code with their fixity rules associated with their symbols. The same is true of C, Java, Python, etc. The only difference is that those languages have a closed set of operators while Haskell's is open. There is a small amount of cognitive overhead, but I think you're greatly exaggerating it. In six and a half years of professional Haskell development I can count on one hand the number of times this has been an issue. And it can be trivially resolved by adding parentheses or by decomposing the expression into separate expressions.