auto keyword in C++

Asked Dec 07 '14 at 13:33

Active Dec 07 '14 at 13:39

Viewed 70 times

Why C++ compiler doesn't guess that its initializer_list and not int?>

#include <iostream>
#include <typeinfo>

int main() {
    auto f = 45; // int, fine.
    auto li {12,2,54,65}; // initializer_list, as expected/
    auto A{12};  // ????
    std::cout<<typeid(A).name()<<std::endl;
    return 0;
}

edited Dec 07 '14 at 13:39

asked Dec 07 '14 at 13:33

Person.Junkie

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What does it say? I'd expect it to be a `signed integer` or not compile at all. – RedX Dec 07 '14 at 13:36
Does that code compile with some compiler; if so, which one? – Cheers and hth. - Alf Dec 07 '14 at 13:36
@sbooth: please refrain from "correcting" the code of a question. that changes the question (unless you really know what you're doing). – Cheers and hth. - Alf Dec 07 '14 at 13:39
sorry for error. EDIT – Person.Junkie Dec 07 '14 at 13:39
@Gui_C: oh well, with latest edit i can report that both mingw g++ and visual c++ deduce the type as `initializer_list`. i wouldn't have expected it to compile. learning something new... – Cheers and hth. - Alf Dec 07 '14 at 13:41
Only N3922 changes the rules making things simpler and more intuitive, so that without equals operator you get `int`, and `auto li{1,2,3}` is a compiler error, which compiler are you using and what is actually your question? – Piotr Skotnicki Dec 07 '14 at 13:43
@Cheersandhth.-Alf I added a missing semicolon, which was clearly an error not related to the question itself. – sbooth Dec 07 '14 at 13:50

auto keyword in C++

0 Answers0