operator-precedence

问题 map<int, int> mp; printf(\"%d \", mp.size()); mp[10]=mp.size(); printf(\"%d\\n\", mp[10]); This code yields an answer that is not very intuitive: 0 1 I understand why it happens - the left side of the assignment returns reference to mp[10] \'s underlying value and at the same time creates aforementioned value, and only then is the right side evaluated, using the newly computed size() of the map. Is this behaviour stated anywhere in C++ standard? Or is the order of evaluation undefined? Result

问题 So, I\'m looking into writing a slightly more complex operation with logic operators in an if-else statement. I know I can do parentheses, and I know it\'s the better way of doing this, but I\'ve gotten curious and so I\'m going to ask. If I were to do something like this: if (firstRun == true || selectedCategory != undefined && selectedState != undefined) { //Do something } else { //Do something else } How will that be operated without the use of parentheses? I know there is an order of

问题 I have a single line code, int a = 10; a = ++a * ( ++a + 5); My expected output was 12 * (11 + 5) = 192 ,but I got 187 . As much as I knew the increment operator inside () is to be solved first, then why the one outside is solved first? 回答1: Expressions are evaluated left to right. Parentheses (and precedence) just express grouping, they don't express ordering of evaluation. So 11 * (12 + 5) ++a ++a equals 187 回答2: Quoting from Eric Lippert's Blog: The evaluation of an arithmetical expression

问题 What would a correct operator precedence table that lists all operators in the C language look like? I have made extensive searches on the web, and found many such precedence tables. Alas, I haven\'t found a single one filling these requirements: Lists all operators in the C language as defined in ISO 9899:2011, without mixing in any C++ operators. Lists the operators in the complete and correct priority order. 回答1: Explanation Prec. denotes operator precedence , where group 1 has the highest

问题 This question already has an answer here: Why are these constructs using pre and post-increment undefined behavior? 14 answers int b=0,a=1;b= ++a + ++a; what is the value of b? what is the calculation for it? [duplicate] 2 answers I\'m new to C language so plz sum1 help me out. A C code written int i=3; printf(\"%d\",++i + ++i); Complier gvs O/P =9. How? Thanx in advance 回答1: The results are undefined. You're modifying a variable more than once in an expression (or sequence point to be more

问题 Why does the following compile in C++? int phew = 53; ++++++++++phew ; The same code fails in C, why? 回答1: That is because in C++ pre-increment operator returns an lvalue and it requires its operand to be an lvalue . ++++++++++phew ; in interpreted as ++(++(++(++(++phew)))) However your code invokes Undefined Behaviour because you are trying to modify the value of phew more than once between two sequence points. In C , pre-increment operator returns an rvalue and requires its operand to be an

问题 I\'ve just started playing around with Scala, and I just learned about how methods can be made right-associative (as opposed to the more traditional left-associativity common in imperative object-oriented languages). At first, when I saw example code to cons a list in Scala, I had noticed that every example always had the List on the right-hand side: println(1 :: List(2, 3, 4)) newList = 42 :: originalList However, even after seeing this over and over again, I didn\'t think twice about it,

问题 In the other topic, @Dietmar gave this solution: template <typename... T> std::tuple<T...> parse(std::istream& in) { return std::tuple<T...>{ T(in)... }; } stating that, The use of brace initialization works because the order of evaluation of the arguments in a brace initializer list is the order in which they appear . (emphasize mine) The relevant text from the C++ Standard (n3485) is, Within the initializer-list of a braced-init-list, the initializer-clauses, including any that result from

问题 I am learning C language and quite confused the differences between ++*ptr and *ptr++ . For example: int x = 19; int *ptr = &x; I know ++*ptr and *ptr++ produce different results but I am not sure why is that? 回答1: These statements produce different results because of the way in which the operators bind. In particular, the prefix ++ operator has the same precedence as * , and they associate right-to-left. Thus ++*ptr is parsed as ++(*ptr) meaning "increment the value pointed at by ptr ,". On

问题 Possible Duplicate: Undefined Behavior and Sequence Points I\'m not sure if this is a gcc bug or not, so I\'ll ask: unsigned int n = 0; std::cout << n++ << n << ++n; gcc gives the extremely strange result: \"122\" which AFAICT is impossible. Because << is left associative, it should be the same as: operator<<(operator<<(operator<<(std::cout, n++), n), ++n) and because there is a sequence point before and after evaluating arguments, n is never modified twice (or even accessed) between two