operator-precedence | 易学教程

Haskell type operator precedence

阅读更多关于 Haskell type operator precedence

问题 When the language extension TypeOperators is enabled, it's possible to define own type operators. Also, it's possible to set their relative precedence with infix* . But what's the precedence of (->) , for example? > :i (->) data (->) a b -- Defined in `GHC.Prim' instance Monad ((->) r) -- Defined in `GHC.Base' instance Functor ((->) r) -- Defined in `GHC.Base' instance Applicative ((->) a) -- Defined in `Control.Applicative' instance Arrow (->) -- Defined in `Control.Arrow' instance Monoid b

Haskell Precedence: Lambda and operator

阅读更多关于 Haskell Precedence: Lambda and operator

问题 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

Pointer operations and operator precedence in C

阅读更多关于 Pointer operations and operator precedence in C

问题 Background Just had a chat with a C guy today and we disagreed on the following: int intgA[2] = { 1, 2 }; int intgB[2] = { 3, 5 }; int *intAPtr = intgA; int *intBPtr = intgB; So when we do: *intAPtr++ = *intBPtr++; My analysis First: intBPtr increments by one, now pointing to the address of 5. Then, deference, holding the value 5; intAPtr also increments by one, now pointing to the address of 2. Subsequently referencing and the value is 2; Lastly: 2 is replaced by 5. So respectively they are:

Haskell infix function application precedence

阅读更多关于 Haskell infix function application precedence

问题 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 precedence of unary operators

阅读更多关于 Operator precedence of unary operators

Some information source on operator precedence like this says that unary operators like ! , ~ , + , - have higher precedence than assignment = . However, the following expressions are possible: !a = true # => false (with warning) a # => true ~a = 1 # => -2 a # => 1 +a = 1 # => 1 a # => 1 -a = 1 # => -1 a # => 1 Considering these results, the only possible explanation I can think of is that these unary operator have lower precedence than the assignment. If that is the case, then it would mean that the information I mentioned above is wrong. Which is correct? Is there a different explanation? My

What is the order of evaluation in python when using pop(), list[-1] and +=?

阅读更多关于 What is the order of evaluation in python when using pop(), list[-1] and +=?

问题 a = [1, 2, 3] a[-1] += a.pop() This results in [1, 6] . a = [1, 2, 3] a[0] += a.pop() This results in [4, 2] . What order of evaluation gives these two results? 回答1: RHS first and then LHS. And at any side, the evaluation order is left to right. a[-1] += a.pop() is same as, a[-1] = a[-1] + a.pop() a = [1,2,3] a[-1] = a[-1] + a.pop() # a = [1, 6] See how the behavior changes when we change the order of the operations at RHS, a = [1,2,3] a[-1] = a.pop() + a[-1] # a = [1, 5] 回答2: The key insight

i = ++i + ++i; in C++

阅读更多关于 i = ++i + ++i; in C++

Can someone explain to me why this code prints 14? I was just asked by another student and couldn't figure it out. int i = 5; i = ++i + ++i; cout<<i; The order of side effects is undefined in C++. Additionally, modifying a variable twice in a single expression has no defined behavior (See the C++ standard , §5.0.4, physical page 87 / logical page 73). Solution: Don't use side effects in complex expression, don't use more than one in simple ones. And it does not hurt to enable all the warnings the compiler can give you: Adding -Wall (gcc) or /Wall /W4 (Visual C++) to the command line yields a

What should be the output of echo ++$a + $a++ [duplicate]

阅读更多关于 What should be the output of echo ++$a + $a++ [duplicate]

This question already has answers here : Why is $a + ++$a == 2? (13 answers) In the PHP manual, operator precedence section , there is this example: // mixing ++ and + produces undefined behavior $a = 1; echo ++$a + $a++; // may print 4 or 5 I understand the behavior is undefined because of the following reason: Since x + y = y + x the interpreter is free to evaluate x and y for addition in any order in order to optimize speed and/or memory. I concluded this after looking at the C code example in this article . My question is that the output of the above mentioned PHP code should be 4 no

When using doubles, why isn't (x / (y * z)) the same as (x / y / z)? [duplicate]

阅读更多关于 When using doubles, why isn't (x / (y * z)) the same as (x / y / z)? [duplicate]

问题 This question already has answers here : How to avoid floating point precision errors with floats or doubles in Java? (12 answers) Double calculation producing odd result [duplicate] (3 answers) Closed 4 years ago . 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() {

showsPrec and operator precedences

阅读更多关于 showsPrec and operator precedences

问题 I asked about this before, but it seems I phrased the question too narrowly. So let's see if I can explain what I'm actually after. Suppose I have some type that supports several binary operators, each with varying precedence and associativity. How do I write a Show instance that correctly brackets sub-expressions? I know I'm being dense here, but I get this wrong every single time I try to do it. There must be some mechanical procedure you can follow to make this work out correctly, but I