This is a long and twisted story. 🙂 Yes, map, set, list, etc. keep counters to supply a constant time size(). Prior to C++11, none of the containers were required to keep counters, as their size() should be constant time, but not shall be constant time. In the standard, “should” means maybe, maybe not, and … Read more
1. gcc builtin function can get the file name of a full path at compile time. #define __FILENAME__ (__builtin_strrchr(__FILE__, “https://stackoverflow.com/”) ? __builtin_strrchr(__FILE__, “https://stackoverflow.com/”) + 1 : __FILE__) or #define __FILENAME__ (strrchr(__FILE__, “https://stackoverflow.com/”) ? strrchr(__FILE__, “https://stackoverflow.com/”) + 1 : __FILE__) 2. c++11 constexpr also can do this at compile time. c++11 constexpr function can only use … Read more
Do rvalue references allow dangling references? If you meant “Is it possible to create dangling rvalue references” then the answer is yes. Your example, however, string middle_name () { return “Jaan”; } int main() { string&& nodanger = middle_name(); // OK. // The life-time of the temporary is extended // to the life-time of the … Read more
You need to set the pre-processor symbol ‘__GXX_EXPERIMENTAL_CXX0X__‘ to the eclipse project. g++ automatically adds that when you use ‘-std=c++0x’, but eclipse is not aware of that, so it treats those sections of the relevant headers as disabled.
Unfortunately, you simply cannot do this! Some static constexpr members may be initialised inline: [C++11 9.4.2/3]: [..] A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. … Read more
If you have reason to suspect there is a worthwhile performance gain to be had, cut it out with the rules of thumb and measure. The purpose of the advise you quote is that you don’t copy great amounts of data for no reason, but don’t jeopardize optimizations by making everything a reference either. If … Read more
Add another pair of braces. std::array<std::pair<int, int>, 2> ids = { { { 0, 1 }, { 1, 2 } } }; std::array<T, N> is an aggregate class containing a member of type T[N]. Usually, you can initialise that the same way you would a plain T[N] array, but when you’re dealing with a non-aggregate … Read more
You cannot use a lambda expression except by actually creating that object- that makes it impossible to pass to type deduction like decltype. Ironically, of course, the lambda return rules make it so that you CAN return lambdas from lambdas, as there are some situations in which the return type doesn’t have to be specified. … Read more
In your case, there is most likely no difference. However, generally, the difference is as follows: using A::foo; resolves A from the current scope, while using ::A::foo searches for A from the root namespace. For example: namespace A { namespace B { class C; } } namespace B { class C; } namespace A { … Read more
What is type of auto in here ? The type is decltype(m_some_class) – I.e., the return value is of the same type as the variable m_some_class. Note that the function will return a copy of *this. If a reference to *this is wanted instead, you can use auto& or, since C++14, the more generic decltype(auto).