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 …
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. …
No. This is not acceptable. This code invokes Undefined behavior. This is because of the operation on b is not defined. In the expression a=(a+b)-(b=a); it is not certain whether b gets modified first or its value gets used in the expression (a+b) because of the lack of the sequence point. See what standard syas: …
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) …
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 …
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 …