operator-precedence

问题 I have run in to some weird thing when I want to print one of my objects (which is obviously not null). If I use this line: text.append("\n [ITEM ID]: " + (item == null ? (otherItem == null ? 0 : otherItem .getItems().get(i).getId()) : item .getItems().get(i).getId())); There is no null pointer exception if my item object is null . Of course this should be the excepted result. But if I use it without the () marks: text.append("\n [ITEM ID]: " + item == null ? (otherItem == null ? 0 :

问题 This question already has answers here : Priority of the logical statements NOT AND & OR in python (5 answers) Closed last year . Which operator takes precedence in 4 > 5 or 3 < 4 and 9 > 8 ? Would this be evaluated to true or false? I know that the statement 3 > 4 or (2 < 3 and 9 > 10) should obviously evaluate to false but I am not quite sure how python would read 4 > 5 or 3 < 4 and 9 > 8 回答1: Comparisons are executed before and , which in turn comes before or . You can look up the

问题 This question already has answers here : Conditional operator used in cout statement (2 answers) Closed 5 years ago . I have the following test program: #include <string> #include <iostream> int main() { std::string s; std::string a = "sd"; std::cout << a==s ? "y" : "n"; return 0; } Trying to compile this with g++ test.cpp gives the following cryptic error: error: no match for 'operator==' (operand types are 'std::basic_ostream<char>' and 'std::string {aka std::basic_string<char>}') std::cout

问题 I had an expression with increment and decrement operators along with some binary operators. public class precedence { public static void main(String ar[]){ int a=3; int b=4; System.out.println(a++ * b-- / a-- + ++b); System.out.println(a+","+b); } } first ++b is replaced by 5 and b will be 5. Then as all remaining terms are post fix versions the order of evaluation will be from right to left a-- will be replaced with 3 and a is changed as 2. b-- will be replaced with 5 and b becomes 4. a++

问题 Example: select count(*) from my table where column1 is not null and (column1 = 4 OR column1 = 5) Example 2: select count(*) from my table where column1 is not null and column1 = 4 OR column1 = 5 In my database with the real column names, I get two different results. The one with the parentheses is right because if I do: select count(*) from my table where column1 is not null and column1 = 4 and then select count(*) from my table where column1 is not null and column1 = 5 and add them together

问题 I have a question with this pointer value assignment: *p++ = *q++; According to Operator Priority Table The priorities of operators are "++" > "*" > "=". But the result of the above statement does the assignment "=" first, as the following *p = *q; p++; q++; Why? 回答1: The post-increment operator increments its operand after its value has already been computed. The pointer dereference therefore occurs on the values the pointers held before this line. However, the precedence you give is correct

问题 I'm reading from a byte array as follows: int* i = (int*)p; id = *i; i++; correct me if I'm wrong, but ++ has precedence over *, so is possible to combine the *i and i++ in the same statement? (e.g. *i++) (this is technically unsafe C#, not C++, p is a byte*) 回答1: I believe that id = *i; i++; and id = *i++; are equivalent. The ++ operator, when used as a suffix (e.g. i++ ), returns the value of the variable prior to the increment. I'm somewhat confused by the reflector output for unsafe class

问题 if an INSERT and a SELECT are done simultaneously on a mysql table which one will go first? Example: Suppose "users" table row count is 0. Then this two queries are run at the same time (assume it's at the same mili/micro second): INSERT into users (id) values (1) and SELECT COUNT(*) from users Will the last query return 0 or 1? 回答1: Depends whether your users table is MyISAM or InnoDB. If it's MyISAM, one statement or the other takes a lock on the table, and there's little you can do to

问题 This leads to the situation like: -1 ** 0.5 #=> -1 Only parenthesis remedies it: (-1) ** 0.5 #=> 6.123031769111886e-17+1.0i which is less favorable then expected 1.i , but basically acceptable. Before I go to Ruby bugs to complain, I would like to know whether there is perhaps some reason for this to be so? 回答1: Many languages define their operator precedence tables by modeling after mathematics' order of operations. In math, exponentiation does have higher precedence than multiplication, and

问题 Please could you explain this apparently inconsistent behaviour to me: use strict; my @a; print "a" x 2; # this prints: aa @a = "a" x 2; print @a; # this prints: aa print ("a") x 2; # this prints: a @a = ("a") x 2; print @a; # this prints: aa Shouldn't the last one print a single 'a'? Edit: Ok so now this is kind of making more sense to me: "Binary "x" is the repetition operator...In list context, if the left operand is enclosed in parentheses or is a list formed by qw/STRING/, it repeats the