Is double-braced scalar initialization allowed by the C++ standard?

up vote
33
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I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

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up vote
33
down vote

favorite

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

add a commentÂ |Â

up vote
33
down vote

favorite

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

I have the following code:

int x = ;

Is this syntax valid according to the C++ standard? (I'm interested in C++11 and later.)

When using the latest compilers there is no problem, however in some older ones (e.g. GCC 4.8.5) it gives following error:

error: braces around scalar initializer for type 'int'

c++ c++11 language-lawyer

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

edited Aug 29 at 13:00

Boann

35.7k1184116

edited Aug 29 at 13:00

Boann

35.7k1184116

edited Aug 29 at 13:00

Boann

35.7k1184116

asked Aug 29 at 10:34

Igor

36939

asked Aug 29 at 10:34

Igor

36939

asked Aug 29 at 10:34

Igor

36939

add a commentÂ |Â

1 Answer
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up vote
34
down vote

accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

If the initializer list is a designated initializer list, [...] => it's not

If T is an aggregate class [...] => it's not

If T is a character array [...] => it's not.

If T is an aggregate [...] => it's not (only arrays and classes are aggregates)

If the initializer list has no elements [...] => it doesn't

If T is a specialization of std::initializer_list [...] => it's not

If T is a class type [...] => it's not

If T is an enumeration with fixed underlying type [...] => it's not

If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no

If T is a reference type [...] => it isn't

If the initializer list has no elements [...] => it doesn't

Otherwise the program is ill-formed

answered Aug 29 at 10:48

Rakete1111

32.7k975110

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

add a commentÂ |Â

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1 Answer
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active

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1 Answer
1

active

oldest

votes

up vote
34
down vote

accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

If the initializer list is a designated initializer list, [...] => it's not

If T is an aggregate class [...] => it's not

If T is a character array [...] => it's not.

If T is an aggregate [...] => it's not (only arrays and classes are aggregates)

If the initializer list has no elements [...] => it doesn't

If T is a specialization of std::initializer_list [...] => it's not

If T is a class type [...] => it's not

If T is an enumeration with fixed underlying type [...] => it's not

If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no

If T is a reference type [...] => it isn't

If the initializer list has no elements [...] => it doesn't

Otherwise the program is ill-formed

answered Aug 29 at 10:48

Rakete1111

32.7k975110

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

add a commentÂ |Â

up vote
34
down vote

accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

If the initializer list is a designated initializer list, [...] => it's not

If T is an aggregate class [...] => it's not

If T is a character array [...] => it's not.

If T is an aggregate [...] => it's not (only arrays and classes are aggregates)

If the initializer list has no elements [...] => it doesn't

If T is a specialization of std::initializer_list [...] => it's not

If T is a class type [...] => it's not

If T is an enumeration with fixed underlying type [...] => it's not

If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no

If T is a reference type [...] => it isn't

If the initializer list has no elements [...] => it doesn't

Otherwise the program is ill-formed

answered Aug 29 at 10:48

Rakete1111

32.7k975110

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

add a commentÂ |Â

up vote
34
down vote

accepted

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

If the initializer list is a designated initializer list, [...] => it's not

If T is an aggregate class [...] => it's not

If T is a character array [...] => it's not.

If T is an aggregate [...] => it's not (only arrays and classes are aggregates)

If the initializer list has no elements [...] => it doesn't

If T is a specialization of std::initializer_list [...] => it's not

If T is a class type [...] => it's not

If T is an enumeration with fixed underlying type [...] => it's not

If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no

If T is a reference type [...] => it isn't

If the initializer list has no elements [...] => it doesn't

Otherwise the program is ill-formed

answered Aug 29 at 10:48

Rakete1111

32.7k975110

This is ill-formed. gcc is wrong to accept it and clang seems to allow it as an extension, as it warns about it.

I'm going to quote the latest draft, but it doesn't make a difference. List initialization works as follows as per [dcl.init.list], where T is int in this case:

If the initializer list is a designated initializer list, [...] => it's not

If T is an aggregate class [...] => it's not

If T is a character array [...] => it's not.

If T is an aggregate [...] => it's not (only arrays and classes are aggregates)

If the initializer list has no elements [...] => it doesn't

If T is a specialization of std::initializer_list [...] => it's not

If T is a class type [...] => it's not

If T is an enumeration with fixed underlying type [...] => it's not

If the initializer list has a single element of type E [...] => a braced initializer list has no type, so no

If T is a reference type [...] => it isn't

If the initializer list has no elements [...] => it doesn't

Otherwise the program is ill-formed

answered Aug 29 at 10:48

Rakete1111

32.7k975110

answered Aug 29 at 10:48

Rakete1111

32.7k975110

answered Aug 29 at 10:48

Rakete1111

32.7k975110

answered Aug 29 at 10:48

Rakete1111

32.7k975110

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

add a commentÂ |Â

4

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

+1 and see also core language issue 1501 which follows the same logic as this answer to explicitly say that it's invalid.
â€“Â hvd
Aug 29 at 11:51

add a commentÂ |Â

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