operator-precedence

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 am reading the source code of Underscore.js, then something confused me: // Its code, check the passed-in parameter obj _.isObject = function(obj) { var type = typeof obj; return type === 'function' || type === 'object' && !!obj; }; I am confused about the operator order of expression. I think the operator precedence in return type === 'function' || type === 'object' && !!obj; will be from left to right ; I mean equal to : return (type === 'function' ) || ( type === 'object' && !!obj); if

问题 I am attempting to perform a numerical optimisation of a "black box" function in Mathematica. Schematically it goes like this: NMinimize[{comb[x,y,z], x > 0}, {x,y,z}] where comb[x,y,z] is defined similarly to this: comb[x_,y_,z_] := Module[{}, Print[x,y,z]; M = FindMaximum[SkewNormal[a,x,y,z], {a,x}] // First; val = f[x,y,z,M]; Return[val]; ]; However, all of the minimisation functions I have tried seem to not immediately provide comb[x,y,z] with numerical values, and it ends up trying to

问题 If I have a piece of code written in C# wrapped in an #if directive, what (if any) precedence is applied to any boolean operators that might be used in that directive? In other words: #if DEBUG || MYTEST && PLATFORM_WINDOWS // ... Some code here #endif Will that be simply evaluated left to right as #if (DEBUG || MYTEST) && PLATFORM_WINDOWS And similarly, would #if PLATFORM_WINDOWS && DEBUG || MYTEST Be evaluated as #if (PLATFORM_WINDOWS && DEBUG) || MYTEST Or is there some precedence order

问题 If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? 回答1: Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]

问题 If I give Mathematica the input TreeForm[Unevaluated[4^5]] I expect to see three boxes -- power, 4, and 5. Instead I see a single box with 1024. Can anyone explain? 回答1: Compare TreeForm@Unevaluated[4^5] with TreeForm@Hold[4^5] From the help: Unevaluated[expr] represents the unevaluated form of expr when it appears as the argument to a function. and Hold[expr] maintains expr in an unevaluated form. so, as Unevaluated[4^5] gets to TreeForm ... it gets evaluated ... It works like this: f[x_+y_]

问题 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. Closed 6 years ago . I have a written a mysql select query to fetch schedule details based on origin states,origin city,destination state and destination city. In my query i

问题 JLS: The lowest precedence operator is the arrow of a lambda expression (->) , followed by the assignment operators. Followed in which direction (increasing priority, decreasing priority)? - "followed" means assignment has higher priority or lower priority (with respect to arrow operator)? I guess, in increasing, because "lowest" (for arrow) means absolutely lowest. As I understand, arrow (->) should be at the very bottom of this Princeton operator precedence table (that is below all

问题 The output of this program: #include <iostream> class c1 { public: c1& meth1(int* ar) { std::cout << "method 1" << std::endl; *ar = 1; return *this; } void meth2(int ar) { std::cout << "method 2:"<< ar << std::endl; } }; int main() { c1 c; int nu = 0; c.meth1(&nu).meth2(nu); } Is: method 1 method 2:0 Why is nu not 1 when meth2() starts? 回答1: Because evaluation order is unspecified. You are seeing nu in main being evaluated to 0 before even meth1 is called. This is the problem with chaining. I

问题 So the operator precedence of the ternary operator in C seems truly bizarre to me. Case in point: #include <stdio.h> int main () { int i=5; int j=6; int k=7; printf("A: %d\n", i+j+(k!=7)?1:11); //prints 1 printf("B: %d\n", i+j+((k!=7)?1:11)); //prints 22 return 0; } This seems similar to the question here: C++ ternary conditional and assignment operator precedence Ternary operator evaluation order As a clarification, I understand that the parentheses make it work, as my comments in my