In class initialization and initializer list

Question

I have recently discovered that you cant have at the same time in class initialization and initializer list. The following code fails :

struct s
{
    int i=0;    
};
int main() {
    s s1; //s1.i = 0
    //s s2={42}; //fails
    return 0;
}

If I remove the in class initialization, the initializer list works fine !

Can someone explains me why a such thing is no allowed ?

Are you sure you're compiling in C++11 mode? Does the compiler really have it implemented? Which compiler and what version of it do you have? — Some programmer dude, Oct 18 '13 at 14:41

bames53 · Accepted Answer · 2013-10-18T15:08:46.990

7

In fact this is allowed in C++14.

struct s
{
    int i=0;    
};

int main() {
    s s1;
    s s2 = {42}; // succeeds
}

It's likely that your compiler just isn't implementing the new rule in C++14. The latest version of clang, however, accepts this and does the correct thing in C++14 mode.

When in-class initialization was added to C++11 it was specified such that it prevented a class from being an aggregate. This was done because at the time the aggregate concept was closely related to PoD types which need to be trivially constructible. Having an in-class initialization means that a type is no longer trivially constructible. Since then, however, the two concepts have become more independent, and so for C++14 a short proposal reversing that decision was accepted.

edited Oct 18 '13 at 15:08

answered Oct 18 '13 at 14:44

bames53

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It doesn't with GCC 4.8. – thokra Oct 18 '13 at 14:46
1

I don't think it is legal in C++11. See my answer. – juanchopanza Oct 18 '13 at 14:46
@thokra I had thought it was in gcc 4.9, but it isn't there either. – bames53 Oct 18 '13 at 14:47
@bames53: Ah, C++14 ... now we're talking. ;) – thokra Oct 18 '13 at 14:48
Oki, thanks for the answer. I'm using g++ 4.8.1 and it dosen't have this feature – Davidbrcz Oct 18 '13 at 14:49
+1 [here is the proposal](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3653.html). – Jesse Good Oct 18 '13 at 14:49
the question is for c++11, where your suggestion doesn't work – BЈовић Oct 18 '13 at 14:50
1

I know it has been selected as the answer and all, but this isn't answering the question. – juanchopanza Oct 18 '13 at 14:52

score 4 · Answer 2 · edited May 23 '17 at 11:45

4

This initialization:

s s1 = { 42 };

requires that s be an aggregate, or that it have a valid constructor taking e.g an int or an std::initializer_list.

When you add a member initialization at the point of declaration, you render your class s a non-aggregate, so you can no longer use aggregate initialization.

You could use the same initialization syntax for your non-aggregate by adding a constructor:

struct s
{
    s(int i) : i(i) {}
    int i=0;    
};

I believe this restriction has been relaxed for C++14.

See What are aggregates... for more information.

edited May 23 '17 at 11:45

Community

1
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answered Oct 18 '13 at 14:42

juanchopanza

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It is not an explanation. The question is why aggregates may not have brace-or equal initializers for their members. – Vlad from Moscow Oct 18 '13 at 14:47
@VladfromMoscow because the standard says so. There isn't any explanation there, but I guess people realised that it was safe enough to do away with this rule for C++14. – juanchopanza Oct 18 '13 at 14:49
I have come to this workaround too but it was precisely what I wanted to avoid – Davidbrcz Oct 18 '13 at 14:50
@CômeDavid I understand. I was not suggesting it as a work-around, just trying to show how a non-aggregate can be constructed with aggregate initialization syntax. – juanchopanza Oct 18 '13 at 14:51

In class initialization and initializer list

2 Answers2