associativity

问题 std::cout << (true ? "high pass" : false ? "fail" : "pass") is the same as std::cout << (true ? "high pass" : (false ? "fail" : "pass")) Since the ternary operator is right associative, why don't we perform the right-hand operation first? Shouldn't pass be printed instead of high pass ? 回答1: You misunderstood operator associativity. It's simply the way to group operators with the same precedence and doesn't affect order of evaluation in any way. So cond1 ? 1 : cond2 ? 2 : cond3 ? 3 : 4 will

问题 Here is a Monad instance for ListT (copied from montrivo) case class ListT[M[_], A](value: M[List[A]]) implicit def listTMonad[M[_]: Monad] = new Monad[ListT[M, *]] { override def flatMap[A, B](fa: ListT[M, A])(f: A => ListT[M, B]): ListT[M, B] = ListT( Monad[M].flatMap[List[A], List[B]](fa.value)( list => Traverse[List].flatTraverse[M, A, B](list)(a => f(a).value) ) ) override def pure[A](a: A): ListT[M, A] = ListT(Monad[M].pure(List(a))) override def tailRecM[A, B](a: A)(f: A => ListT[M,

问题 Here is a Monad instance for ListT (copied from montrivo) case class ListT[M[_], A](value: M[List[A]]) implicit def listTMonad[M[_]: Monad] = new Monad[ListT[M, *]] { override def flatMap[A, B](fa: ListT[M, A])(f: A => ListT[M, B]): ListT[M, B] = ListT( Monad[M].flatMap[List[A], List[B]](fa.value)( list => Traverse[List].flatTraverse[M, A, B](list)(a => f(a).value) ) ) override def pure[A](a: A): ListT[M, A] = ListT(Monad[M].pure(List(a))) override def tailRecM[A, B](a: A)(f: A => ListT[M,

问题 In any programming language textbooks, we are always told how each operator in that language has either left or right associativity. It seems that associativity is a fundamental property of any operator regardless of the number of operands it takes. It also seems to me that we can assign any associativity to any operator regardless of how we assign associativity to other operators. But why is it the case? Perhaps an example is better. Suppose I want to design a hypothetical programming

问题 In any programming language textbooks, we are always told how each operator in that language has either left or right associativity. It seems that associativity is a fundamental property of any operator regardless of the number of operands it takes. It also seems to me that we can assign any associativity to any operator regardless of how we assign associativity to other operators. But why is it the case? Perhaps an example is better. Suppose I want to design a hypothetical programming

问题 I'm trying to prove that type-level function Union is associative, but I'm not sure how it should be done. I proved right identity and associativity laws for type-level append function and right identity for union: data SList (i :: [k]) where SNil :: SList '[] SSucc :: SList t -> SList (h ': t) appRightId :: SList xs -> xs :~: (xs :++ '[]) appRightId SNil = Refl appRightId (SSucc xs) = case appRightId xs of Refl -> Refl appAssoc :: SList xs -> Proxy ys -> Proxy zs -> (xs :++ (ys :++ zs)) :~:

问题 It was very hard to come up with a title... (I'm not a native English speaker.) struct A { int value; A operator+(int i) const { A a; a.value=value+i; return a; }; }; int main(){ A a; a.value=2; a=a+2; return 0; } This code compiles/works as expected, but when I change a=a+2 to a=2+a, it won't compile anymore. GCC gives me this error: no match for ”operator+” in ”2 + a” Is there any way to somehow make 2+a work just like a+2? 回答1: You need a free function, defined after the class struct A { /

问题 Introduction In every textbook on C/C++, you'll find an operator precedence and associativity table such as the following: http://en.cppreference.com/w/cpp/language/operator_precedence One of the questions on StackOverflow asked something like this: What order do the following functions execute: f1() * f2() + f3(); f1() + f2() * f3(); Referring to the previous chart I confidently replied that functions have left-to-right associativity so in the previous statements the are evaluated like this

问题 Example: select count(*) from my table where column1 is not null and (column1 = 4 OR column1 = 5) Example 2: select count(*) from my table where column1 is not null and column1 = 4 OR column1 = 5 In my database with the real column names, I get two different results. The one with the parentheses is right because if I do: select count(*) from my table where column1 is not null and column1 = 4 and then select count(*) from my table where column1 is not null and column1 = 5 and add them together

问题 I am having trouble implementing a function that given something like this: '((+(d + e) + f) + (a + (a + c)) Returns this: '(d + (e + (f + (a + (a + c))))) Now the catch is that I have to use associativity to get the answer so I only want to use this property: '((a + b) + c) = '(a + (b + c)) To get the correct output. I know how to implement the function for this case: '((a + b) + c) -> '(a + (b + c)) However I cannot seem to figure out how to implement it for the first case above (or any