operator-precedence

问题 Consider the following example: class Quirky { public static void main(String[] args) { int x = 1; int y = 3; System.out.println(x == (x = y)); // false x = 1; // reset System.out.println((x = y) == x); // true } } I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side ( x = y ) which, by the order implied by brackets, should be calculated first. Why does the first expression evaluate to

问题 In C++: assert( std::is_same<int , int>::value ); // does not compile assert( (std::is_same<int , int>::value) ); // compiles Can anyone explain why? 回答1: The comma is being treated as a argument separator for the macro, but parenthesis in your second case protect the arguments. We can see this by going to the draft C++ standard section 16.3 Macro replacement which says ( emphasis mine ): The sequence of preprocessing tokens bounded by the outside-most matching parentheses forms the list of

问题 This is just fine: def foo a or b end This is also fine: def foo return a || b end This returns void value expression : def foo return a or b end Why? It doesn't even get executed; it fails the syntax check. What does void value expression mean? 回答1: return a or b is interpreted as (return a) or b , and so the value of return a is necessary to calculate the value of (return a) or b , but since return never leaves a value in place (because it escapes from that position), it is not designed to

问题 This is just fine: def foo a or b end This is also fine: def foo return a || b end This returns void value expression : def foo return a or b end Why? It doesn't even get executed; it fails the syntax check. What does void value expression mean? 回答1: return a or b is interpreted as (return a) or b , and so the value of return a is necessary to calculate the value of (return a) or b , but since return never leaves a value in place (because it escapes from that position), it is not designed to

问题 This question already has answers here : Is Python's order of evaluation of function arguments and operands deterministic (+ where is it documented)? (2 answers) Closed 2 years ago . Is it safe to assume that function arguments are evaluated from left to right in Python? Reference states that it happens that way but perhaps there is some way to change this order which may break my code. What I want to do is to add time stamp for function call: l = [] l.append(f(), time.time()) I understand

问题 I wanted to know how PHP would execute this. Order of operations addslashes(strip_tags($record['value'])); Is addslashes called first or strip_tags? In other words, does it execute from the inside out or from the outside in? 回答1: From the inside out. The things passed into a function in PHP are called "expressions". When you pass an expression as a parameter, what you're really passing is the value of that expression. In order to do that, the expression is evaluated before it is passed in.

问题 This is just fine: def foo a or b end This is also fine: def foo return a || b end This returns void value expression : def foo return a or b end Why? It doesn't even get executed; it fails the syntax check. What does void value expression mean? 回答1: return a or b is interpreted as (return a) or b , and so the value of return a is necessary to calculate the value of (return a) or b , but since return never leaves a value in place (because it escapes from that position), it is not designed to

问题 This is just fine: def foo a or b end This is also fine: def foo return a || b end This returns void value expression : def foo return a or b end Why? It doesn't even get executed; it fails the syntax check. What does void value expression mean? 回答1: return a or b is interpreted as (return a) or b , and so the value of return a is necessary to calculate the value of (return a) or b , but since return never leaves a value in place (because it escapes from that position), it is not designed to

问题 Application has higher precedence than abstraction. In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over? 回答1: Lambda abstraction is λx.M , for some variable x and arbitrary term M . Application is (MN) , for some arbitrary terms M and N . Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic,

问题 Application has higher precedence than abstraction. In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over? 回答1: Lambda abstraction is λx.M , for some variable x and arbitrary term M . Application is (MN) , for some arbitrary terms M and N . Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic,