See the spec, in the compatibility appendix C.1.6: 7.1.6 [see also 3.5] Change: const objects must be initialized in C++ but can be left uninitialized in C Rationale: A const object cannot be assigned to so it must be initialized to hold a useful value. Effect on original feature: Deletion of semantically well-defined feature. Difficulty … Read more
The attempt to write to *dst is UB if the caller passes you a pointer to a const object, or a pointer to a string literal. But if the caller passes you a pointer to data that in fact is mutable, then behavior is defined. Creating a const char* that points to a modifiable char … Read more
I’ll answer your question… the other way around. Why did they allowed Foo const& foo = fooByValue(); to begin with ? It makes life (somewhat) easier, but introduces potential undefined behavior all over the place. Foo const& fooByReference() { return fooByValue(); // error: returning a reference to a temporary } This is obviously wrong, and … Read more
C++ FAQ: If you find ordinary type safety helps you get systems correct (it does; especially in large systems), you’ll find const correctness helps also. You should use const when you want to be sure not to change variable accidentally or intentionally. Some constants (globals and class static, strings & integers, but not variables with … Read more
C99 ยง6.7.3/4: If the same qualifier appears more than once in the same specifier-qualifier-list, either directly or via one or more typedef s, the behavior is the same as if it appeared only once. Yes, that is valid C99, and your discovery is correct.
Sorry, that’s not the way constants in PHP work. You can put variables in double quotes and heredocs but not constants.
Constants defined using define() are fairly slow in PHP. People actually wrote extensions (like hidef) to improve the performance. But unless you have loads of constants this shouldn’t make much of a difference. As of PHP 5.3 you can also use compile-time constants using const NAME = VALUE;. Those are much faster.
I’ve added a few lines to your code. That’s sufficient to make it clear why this is disallowed: void f(vector<const T*>& p) { static const T ct; p.push_back(&ct); // adds a const T* to nonConstVec ! } int main() { vector<T*> nonConstVec; f(nonConstVec); nonConstVec.back()->nonConstFunction(); }
Because you are not changing any variable in X. Actually, you are changing _y which is an outsider with respect to your class. Don’t forget that: y = newY; Is assigning the value of newY to the variable pointed by y, but not the references them selves. Only on initialization the references are considered.
This public final static long TESTNG_TEST_TIMEOUT = 300000; is a constant variable, a type of constant expression. This public final static long TESTNG_TEST_TIMEOUT = TimeUnit.MINUTES.toMillis(5); is not. Annotation members expect constant expressions (and a few other things like enums and Class literals).