I believe behavior is well-defined because of 8[dcl.decl]/3 Each init-declarator in a declaration is analyzed separately as if it was in a declaration by itself. Which is even additionally explained in a footnote as A declaration with several declarators is usually equivalent to the corresponding sequence of declarations each with a single declarator. That is … Read more
My question are; 1.What does it mean by, if an object is written to within a full expression, any and all accesses to it within the same expression must be directly involved in the computation of the value to be written.? With a sub-expression like i++, i is written to. Moreover, assignment is an expression, … Read more
There is a difference between an expression being evaluated and completing its side effects. The b = a assignment expression will be evaluated ahead of subtraction due to higher precedence of the parentheses. It will provide the value of a as the result of the evaluation. The writing of that value into b, however, may … Read more
The expression f() + g() contains a minimum of 4 sequence points; one before the call to f() (after all zero of its arguments are evaluated); one before the call to g() (after all zero of its arguments are evaluated); one as the call to f() returns; and one as the call to g() returns. … Read more
Native operator expressions are not equivalent to overloaded operator expressions. There is a sequence point at the binding of values to function arguments, which makes the operator++() versions well-defined. But that doesn’t exist for the native-type case. In all four cases, i changes twice within the full-expression. Since no ,, ||, or && appear in … Read more
This question gets asked a fair amount. Be warned that every time someone asks this question a great many people post incorrect answers. Lots of people have incorrect ideas about how these operators work, including people who write programming books and thereby teach other people falsehoods. Read the other answers here very carefully. For a … Read more
It looks like the code i.operator+=(i.operator ++()); Works perfectly fine with regards to sequence points. Section 1.9.17 of the C++ ISO standard says this about sequence points and function evaluation: When calling a function (whether or not the function is inline), there is a sequence point after the evaluation of all function arguments (if any) … Read more
When a sequence point occurs, it basically means that you are guaranteed that all previous operations are complete. Changing a variable twice without an intervening sequence point is one example of undefined behaviour. For example, i = i++; is undefined because there’s no sequence point between the two changes to i. Note that it’s not … Read more
C++98 and C++03 This answer is for the older versions of the C++ standard. The C++11 and C++14 versions of the standard do not formally contain ‘sequence points’; operations are ‘sequenced before’ or ‘unsequenced’ or ‘indeterminately sequenced’ instead. The net effect is essentially the same, but the terminology is different. Disclaimer : Okay. This answer … Read more
C has the concept of undefined behavior, i.e. some language constructs are syntactically valid but you can’t predict the behavior when the code is run. As far as I know, the standard doesn’t explicitly say why the concept of undefined behavior exists. In my mind, it’s simply because the language designers wanted there to be … Read more