operator-precedence

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

问题 It seems this question only answered for Java but I would like to know how it works in Python. So are these the same? a += b / 2 and a += (b / 2) 回答1: Yes, those are the same. Python's augmented assignment is not an expression , it is a statement, and doesn't play in expression precedence rules. += is not an operator, and instead it's part of the augmented assignment statement syntax. So everything to the right of the += is an expression, but += itself is not, so the assignment will always be

问题 So I was fooling around with the new exponentiation operator and I discovered you cannot put a unary operator immediately before the base number. let result = -2 ** 2; // syntax error let result = -(2 ** 2); // -4 let x = 3; let result = --x ** 2; // 4 From the documentation on MDN: In JavaScript, it is impossible to write an ambiguous exponentiation expression, i.e. you cannot put a unary operator ( + / - / ~ / ! / delete / void / typeof ) immediately before the base number. In most

问题 I've noticed some scripts seem to be called before others on a certain page, I was wondering, what is the specific order for scripts to load? In-page before referenced .js scripts? Are they run in order from first <script> mentioned to last in page, or Is this browser-dependent? How can one make sure that a specific script is first to run in a page? 回答1: I've noticed some scripts seem to be called before others on a certain page. I was wondering, what is the specific order for scripts to load

问题 So I know that C++ has an Operator Precedence and that int x = ++i + i++; is undefined because pre++ and post++ are at the same level and thus there is no way to tell which one will get calculated first. But what I was wondering is if int i = 1/2/3; is undefined. The reason I ask is because there are multiple ways to look at that (1/2)/3 OR 1/(2/3). My guess is that it is a undefined behavior but I would like to confirm it. 回答1: In your example the compiler is free to evaluate "1" "2" and "3"

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: C#: Order of function evaluation (vs C) Code snippet: i += ++i; a[++i] = i; int result = fun() - gun(); //statement of similar kind Are their behavior well-defined in C#? In C++, such code invokes undefined/unspecified behavior. Please also quote the relevant sections from the language specification in your response! 回答1: The key here is the table in "1.4 Expressions", and "7.3.1 Operator precedence and

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 I recently came along this method for swapping the values of two variables without using a third variable. a^=b^=a^=b But when I tried the above code on different compilers, I got different results, some gave correct results, some didn't. Is anything terribly wrong with the code? 回答1: Is anything terribly wrong with the code? Yes! a^=b^=a^=b in fact invokes Undefined Behaviour in C and in C++ because you are trying to change the value of a more than once between two sequence points. Try

问题 Suppose I do setTimeout(foo, 0); ... setTimeout(bar, 0); Can I be sure foo will begin executing before bar? What if instead of 0 I use a timeout of 1, 10, or 100 for bar? Simple experiments show that in the case of equal timeout values the timeout targets are executed in the same order as the setTimeouts themselves, but is it safe to rely on this behavior? 回答1: It is not safe to rely on this behavior. I wrote a test script which schedules a number of functions (and they in turn also schedule

问题 What's happening here? #include <iostream> using namespace std; int main(){ int x=0,y=0; true? ++x, ++y : --x, --y; cout << "x: " << x << endl; cout << "y: " << y << endl; //why does y=0 here? x=0,y=0; false ? ++x, ++y : --x, --y; cout << "x: " << x << endl; cout << "y: " << y << endl; } x: 1 y: 0 x: -1 y: -1 The second case seems fine. I would expect both x and y to increment to 1 in the first case but only the left hand operand increments. 回答1: The first one is equivalent to: (true ? (++x,