operator-precedence

I found precedence and associativity is a big obstacle for me to understand what the grammar is trying to express at first glance to haskell code. For example, blockyPlain :: Monad m => m t -> m t1 -> m (t, t1) blockyPlain xs ys = xs >>= \x -> ys >>= \y -> return (x, y) By experiment, I finally got it means, blockyPlain xs ys = xs >>= (\x -> (ys >>= (\y -> return (x, y)))) instead of blockyPlain xs ys = xs >>= (\x -> ys) >>= (\y -> return (x, y)) Which works as: *Main> blockyPlain [1,2,3] [4,5,6] [(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)] I can get info from ghci for (>>=) as an

Let f x y = x * y . We can apply this function in two ways: f 5 6 , or, using infix notation, 5 `f` 6 . Do the operator rules apply to this last expression? What precedence will this application have? Is it just another form of function application, and so will it also have the highest precedence? I suppose that the compiler sees this special form (due to `` and/or the name starting with a letter(?)), and actually treats this as ordinary function application, instead of considering it an operator. The Haskell 98 Report has a section on Operator Applications that clears it up: An operator is

I recently came across something that I thought I understood right off the bat, but thinking more on it I would like understanding on why it works the way it does. Consider the code below. The (x-- == 9) is clearly getting evaluated, while the (y++ == 11) is not. My first thought was that logical && kicks in, sees that the expression has already become false, and kicks out before evaluating the second part of the expression. The more I think about it, the more I don't understand why this behaves as it does. As I understand it, logical operators fall below increment operations in the order of

This snippet of Perl code in my program is giving the wrong result. $condition ? $a = 2 : $a = 3 ; print $a; No matter what the value of $condition is, the output is always 3, how come? This is explained in the Perl documentation . Because of Perl operator precedence the statement is being parsed as ($condition ? $a= 2 : $a ) = 3 ; Because the ?: operator produces an assignable result, 3 is assigned to the result of the condition. When $condition is true this means ($a=2)=3 giving $a=3 When $condition is false this means ($a)=3 giving $a=3 The correct way to write this is $a = ( $condition ? 2

This question already has an answer here: How to avoid floating point precision errors with floats or doubles in Java? 12 answers Double calculation producing odd result [duplicate] 3 answers This is partly academic, as for my purposes I only need it rounded to two decimal places; but I am keen to know what is going on to produce two slightly different results. This is the test that I wrote to narrow it to the simplest implementation: @Test public void shouldEqual() { double expected = 450.00d / (7d * 60); // 1.0714285714285714 double actual = 450.00d / 7d / 60; // 1.0714285714285716

This (note the comma operator ): #include <iostream> int main() { int x; x = 2, 3; std::cout << x << "\n"; return 0; } outputs 2 . However, if you use return with the comma operator, this: #include <iostream> int f() { return 2, 3; } int main() { int x; x = f(); std::cout << x << "\n"; return 0; } outputs 3 . Why is the comma operator behaving differently with return ? According to the Operator Precedence , comma operator has lower precedence than operator= , so x = 2,3; is equivalent to (x = 2),3; . (Operator precedence determines how operator will be bound to its arguments, tighter or looser

I followed the explanation given in the "Precedence climbing" section on this webpage to implement an arithmetic evaluator using the precedence climbing algorithm with various unary prefix and binary infix operators. I would also like to include ternary operators (namely the ternary conditional operator ?: ). The algorithm given on the webpage uses the following grammar: E --> Exp(0) Exp(p) --> P {B Exp(q)} P --> U Exp(q) | "(" E ")" | v B --> "+" | "-" | "*" |"/" | "^" | "||" | "&&" | "=" U --> "-" How can I incorporate ternary operators into this grammar? To be specific, I'll use C/C++/Java