operator-precedence

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. Explanation Prec. denotes operator precedence , where group 1 has the highest precedence and group 17 the lowest. Assoc. denotes operator associativity , where such is applicable.

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 The results are undefined. You're modifying a variable more than once in an expression (or sequence point to be more accurate). Modifying a variable more than once between sequence points is undefined, so don't do it. It might be

I dont understand how that output (" four ") comes? $a = 2; echo $a == 1 ? 'one' : $a == 2 ? 'two' : $a == 3 ? 'three' : $a == 5 ? 'four' : 'other' ; // prints 'four' I don't understand why " four " gets printed. You need to bracket the ternary conditionals: <?php for ($a=0; $a < 7; $a++) { echo ( $a == 1 ? 'one' : ($a == 2 ? 'two' : ($a == 3 ? 'three' : ($a == 5 ? 'four' : 'other')))); echo "\n"; // prints 'four' } exit; ?> returns: other one two three other four other as you'd expect. See the note at the bottom of "Ternary operators" at PHP Ternary operator help . The expressions are being

Why does the following compile in C++? int phew = 53; ++++++++++phew ; The same code fails in C, why? Prasoon Saurav 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 lvalue . So your code doesn't compile in C mode. Note: The two defect reports DR#637 and DR

Suppose I have a number of statements that I want to execute in a fixed order. I want to use g++ with optimization level 2, so some statements could be reordered. What tools does one have to enforce a certain ordering of statements? Consider the following example. using Clock = std::chrono::high_resolution_clock; auto t1 = Clock::now(); // Statement 1 foo(); // Statement 2 auto t2 = Clock::now(); // Statement 3 auto elapsedTime = t2 - t1; In this example it is important that the statements 1-3 are executed in the given order. However, can't the compiler think statement 2 is independent of 1

The Python documentation for operator precedence states: Operators in the same box group left to right (except for comparisons, including tests, which all have the same precedence and chain from left to right — see section Comparisons ...) What does this mean? Specifically: "Operators in the same box group left to right (except for comparisons...)" -- do comparisons not group left to right? If comparisons do not group left to right, what do they do instead? Do they "chain" as opposed to "group"? If comparisons "chain" rather than "group", what is the difference between "chaining" and "grouping

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 pack expansions (14.5.3), are evaluated in the order in which they appear. That is, every value

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? templatetypedef 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 the other hand, the postfix ++ operator has higher precedence than the dereferrence operator