std::vector with function-local type

Question

Why does this

#include <vector>

void f()
{
    struct S
    {
        int first, second, third;
    };

    std::vector<S> vs;
}

work with Visual C++ 2015, but not with g++ 4.8.4?

[I don't get any errors here](https://godbolt.org/g/A9igjx). Please explain how it does not work. — NathanOliver, Feb 07 '17 at 18:24
Are you passing `-std=c++0x`? C++11 is enabled by default in MSVC, but not in GCC. (Assuming GCC 4.8 has support for this) — peppe, Feb 07 '17 at 18:24
@NathanOliver [I do](https://godbolt.org/g/16Y6jE), it's missing C++11 flag. Not exactly sure why? — Borgleader, Feb 07 '17 at 18:25
@Borgleader That is interesting. If you go up to version 6.1 then it works again. — NathanOliver, Feb 07 '17 at 18:28
Because GCC 6 defaults to C++14: https://gcc.gnu.org/gcc-6/changes.html "The default mode has been changed to -std=gnu++14." — peppe, Feb 07 '17 at 18:29

score 8 · Answer 1 · edited May 23 '17 at 11:45

8

Ensure you are compiling with at least -std=c++0x.

In C++11, the standard was modified to allow for local classes to be template arguments (for lambda support). If you target pre-C++11, this won't work.

If you're compiling MSVC, it will enable C++11 by default, which is not so with clang and pre-gcc 6

See also: What restrictions does ISO C++03 place on structs defined at function scope?

Demo (GCC 4.8.4) w/ C++11 (Works)
Demo (GCC 4.8.4) without C++11 (Doesn't work)

edited May 23 '17 at 11:45

Community

1
1

answered Feb 07 '17 at 18:27

AndyG

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Might be worth mentionning that this support is on by default in MSVC? (It's been mentioned in the comments but still) – Borgleader Feb 07 '17 at 18:28
"classes to be local to functions" That was allowed in C++98. What was not allowed was to use them as template type parameters . – peppe Feb 07 '17 at 18:29
I was going to ask why C++11 allows local structs. Didnt know that before. When you say "for lambda support" is it something going on behind the scenes when I declare a lambda or something supposed to be used with lambdas? – 463035818_is_not_an_ai Feb 07 '17 at 18:36
4

@tobi303 A lambda is a unnamed class type. It looks like an expression but that is just syntactic sugar. The compiler actually builds a class type with what you provide in the expression. – NathanOliver Feb 07 '17 at 18:37
1

@NathanOliver: And to be slightly more formal, the standard requires that a lambda be a class "The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion class type" [expr.prim.lambda] – AndyG Feb 07 '17 at 18:40
@NathanOliver thx. So a lambda is kind of the same thing as a local functor – 463035818_is_not_an_ai Feb 07 '17 at 18:40
@tobi303 Exactly. It is just a nice short way to be able to write it. Think of them as throw away objects basically. You really should only be using them for something where a named object really isn't need but a object would be nice to have. – NathanOliver Feb 07 '17 at 18:42
That doesn't quite explain what lambdas have to do with allowing local classes as template type parameters... – peppe Feb 07 '17 at 19:16
1

@peppe: Consider `std::copy_if` - its predicate is a template type parameter, and a typical use case of a lambda. – MSalters Feb 08 '17 at 08:55

std::vector with function-local type

1 Answers1