operator-precedence

I read about python following PEMDAS that is precedence of multiply is more than division. I ran the following script print 6*2/1*2 Thus python should interpret this like 12/2 i.e 6 , since precedence of multiplication is more than division. But, the answer is 24. Could anyone let me know where the problem is? Thanks! * has the same operator precedence as / . Operators in the same group evaluate left to right, so your expression evaluates as: 6*2 = 12 / 1 = 12 * 2 = 24 Order of precedence in Python P E M D Left to right A S Left to right 来源： https://stackoverflow.com/questions/35845566

I understand that there are no guarantees regarding the order which the parameters of a function will be called, but, isn't it guaranteed that if there's a function call as parameter, that function will be called first? I help students on the laboratory of an introductory subject on programming, and they were supposed to create a recursive factorial function which receives the n (for n!) and a pointer to an integer that will be used to count function calls and then they are supposed to print the results (n, n! and count). Many were complaining that their use of pointers were wrong so I looked

Possible Duplicate: Problem with PHP ternary operator I was reading up a bit on PHP in this article , and I stopped for a while to consider one of his gripes. I can't figure out how on earth PHP comes to the result that it does. Unlike (literally!) every other language with a similar operator, ?: is left associative. So this: $arg = 'T'; $vehicle = ( ( $arg == 'B' ) ? 'bus' : ( $arg == 'A' ) ? 'airplane' : ( $arg == 'T' ) ? 'train' : ( $arg == 'C' ) ? 'car' : ( $arg == 'H' ) ? 'horse' : 'feet' ); echo $vehicle; prints horse. What logical path does PHP follow that results in 'horse' being

Given std::vector<CMyClass> objects; CMyClass list[MAX_OBJECT_COUNT]; Is it wise to do this? for(unsigned int i = 0; i < objects.size(); list[i] = objects.at(i++)); Or should I expand my loop to this? for(unsigned int i = 0; i < objects.size(); i++) { list[i] = objects.at(i); } The former is undefined behavior. It's not specified whether list[i] is evaluated (to provide an lvalue for the lhs of the assignment) before or after the function call to objects.at . Hence there is a legal ordering of the various parts of the expression, in which i is accessed (in list[i] ) and separately modified (in

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 languages like PHP and Python and others that have an exponentiation operator (typically ^ or ** ), the

I sometimes find this inconvenient when doing bit manipulation (though I can't recall to mind any specific examples right now). I also find it conceptually confusing, since shifts are basically multiplication and division by powers of two. I see that it can be convenient in C++, when using << to send output to an ostream, but of course that can't be used to explain how the order was originally fixed in C. Because that's what the authors of the C language decided. Use parentheses to avoid confusion. 来源： https://stackoverflow.com/questions/9370844/why-do-shifts-have-lower-precedence-than

I get the following: puts true or true and false # >> true whereas I also get: if true or true and false puts "that's true!" else puts "that's false!" end # >> that's false! Why is true or true and false both true and false (like Schrödinger's cat)? It has to do with precedence. puts true or true and false actually evaluates as (puts true) or (true and false) [EDIT: Not exactly. See the note from Todd below.] , and if true or true and false evaluates as if (true or (true and false)) . This is due to the precedences of puts (a method) and if (a language keyword) relative to the other terms of

if (typeof foo !== 'undefined') { // Now we know that foo is defined, we are good to go. } The typeof evaluates to true or false based on whether the variable foo is defined or not. But, say if foo !== 'undefined' evaluates to true , then typeof of true should evaluate to 'boolean' . Why does it evaluate to true or false ? Because precedence rules for the typeof and inquality operators define that that expression is parsed as (typeof foo) !== 'undefined' For more information, see the MDN page on operator precedence . typeof is precedence 16; inquality is precedence 10. The higher precedence of

I have the following code in production that appears to be causing an infinite loop. $z=1; while (!$apns = $this->getApns($streamContext) && $z < 11) { myerror_log("unable to conncect to apple. sleep for 2 seconds and try again"); $z++; sleep(2); } How are the precedence rules getting applied that cause this behavior? http://php.net/manual/en/language.operators.precedence.php I see this note in the docs: Although = has a lower precedence than most other operators, PHP will still allow expressions similar to the following: if (!$a = foo()), in which case the return value of foo() is put into $a