temporary

If I want to get the Cartesian Product of these two vector<string> s: vector<string> final{"a","b","c"}; vector<string> temp{"1","2"}; But I want to put the result in final , such that final would contain: a1 a2 b1 b2 c1 c2 I'd like to do this without creating a temporary array. Is it possible to do this? If it matters, the order of final is not important. You may try the following approach #include <iostream> #include <vector> #include <string> int main() { std::vector<std::string> final{ "a", "b", "c" }; std::vector<std::string> temp{ "1", "2" }; auto n = final.size(); final.resize( final

Possible Duplicate: Binding temporary to a lvalue reference With VS2008 C++ compiler, the codes are compiled without compile error. class A{}; int main(){ A& a_ref = A(); return 0; } I believe the C++ standard, both C++2003 and coming C++0x, disallow it. And I also get a compile time error with gcc compiler. So what I want to know is, is this a known bug for VS compiler to allow initializing reference from a temporary object. Or is it a feature extension of VS compiler? If yes, what's life cycle of the temporary object? It is the extension.This link explains it. What if we take out the const

In MySQL, can temporary variables be used in the WHERE clause? For example, in the following query: SELECT `id`, @var := `id` * 2 FROM `user` @var is successfully set to twice the value of `id` However, if I try to filter the result set to only include results where @var is less than 10: SELECT `id`, @var := `id` * 2 FROM `user` WHERE @var < 10 then I get no results. How can I filter the results based on the value of @var? You need to assign an alias, and test it in the HAVING clause: SELECT id, @var := id * 2 AS id_times_2 FROM user HAVING id_times_2 < 10 Note that if you're just using the

The code below illustrated my concern: #include <iostream> struct O { ~O() { std::cout << "~O()\n"; } }; struct wrapper { O const& val; ~wrapper() { std::cout << "~wrapper()\n"; } }; struct wrapperEx // with explicit ctor { O const& val; explicit wrapperEx(O const& val) : val(val) {} ~wrapperEx() { std::cout << "~wrapperEx()\n"; } }; template<class T> T&& f(T&& t) { return std::forward<T>(t); } int main() { std::cout << "case 1-----------\n"; { auto&& a = wrapper{O()}; std::cout << "end-scope\n"; } std::cout << "case 2-----------\n"; { auto a = wrapper{O()}; std::cout << "end-scope\n"; } std:

I always believed that temporary objects in C++ are automatically considered as const by the compiler. But recently I experienced that the following example of code: function_returning_object().some_non_const_method(); is valid for C++ compiler. And it makes me wonder - are temporary objects in C++ const indeed? If yes, then why the code above is considered correct by the compiler? No, they're not. Not unless you declare the return type as const. It depends. int f(); const int g(); class C { }; C x(); const C y(); In the case of both f() and g() , the returned value is not const because there