operator-precedence

问题 after reading about sequence points, I learned that i = ++i is undefined. So how about this code: int i; int *p = &i; int *q = &i; *p = ++(*q); // that should also be undefined right? Let's say if initialization of p and q depends on some (complicated) condition. And they may be pointing to same object like in above case. What will happen? If it is undefined, what tools can we use to detect? Edit: If two pointers are not supposed to point to same object, can we use C99 restrict? Is it what

问题 Can someone explain to me why this code prints 14? I was just asked by another student and couldn't figure it out. int i = 5; i = ++i + ++i; cout<<i; 回答1: The order of side effects is undefined in C++. Additionally, modifying a variable twice in a single expression has no defined behavior (See the C++ standard, §5.0.4, physical page 87 / logical page 73). Solution: Don't use side effects in complex expression, don't use more than one in simple ones. And it does not hurt to enable all the

问题 I've seen many manifestations of ways to use the user class path as precedent to the hadoop one. Often times this is done if an m/r job needs a specific version of a library that hadoop coincidentally already uses an older version of (for example jackson's json parser or commons http , etc.) In any case : I've seen : mapreduce.task.classpath.user.precedence mapreduce.task.classpath.first mapreduce.job.user.classpath.first Which one of these parameters is the right one to set in my job

问题 This (note the comma operator ): #include <iostream> int main() { int x; x = 2, 3; std::cout << x << "\n"; return 0; } outputs 2 . However, if you use return with the comma operator, this: #include <iostream> int f() { return 2, 3; } int main() { int x; x = f(); std::cout << x << "\n"; return 0; } outputs 3 . Why is the comma operator behaving differently with return ? 回答1: According to the Operator Precedence, comma operator has lower precedence than operator= , so x = 2,3; is equivalent to

问题 I couldn't figure out the following behaviour, String str1= "abc"; String str2 = "abc"; System.out.println("str1==str2 "+ str1==str2); System.out.println("str1==str2 " + (str1==str2)) Output for the above statement is as follows: false str1==str2 true Why is this happening? Why the output is not like follows: str1==str2 true str1==str2 true 回答1: + has higher precedence than ==. So your code : System.out.println("str1==str2 " + str1 == str2); will effectively be System.out.println(("str1==str2

问题 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)? 回答1: 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)) .

问题 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 ? 回答1: 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

问题 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. 回答1: Because that's what the authors of the C language decided. Use parentheses to avoid

问题 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

问题 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