enum{f,g=f};

[Francois Grieu]

Hi, is this valid?

enum{f,g=f};

The problem (if any) is that the value for g references f,
which is defined in an enum not yet completed.

TIA,
Francois Grieu

 

[John]

Hi, is this valid?

enum{f,g=f};

The problem (if any) is that the value for g references f,
which is defined in an enum not yet completed.

TIA,
   Francois Grieu

f == 0
g == 0

 

[Keith Thompson]

Hi, is this valid?

enum{f,g=f};

The problem (if any) is that the value for g references f,
which is defined in an enum not yet completed.

Yes, it's valid.

C99 6.2.1p7:

Each enumeration constant has scope that begins just after the
appearance of its defining enumerator in an enumerator list.

Note that an "enumerator" includes the "= constant" if present, so
"enum { x = x }" is illegal (unless another x was previously declared
in an enclosing scope, but that's ugly so don't do it).

 

[Francois Grieu]

Keith Thompson wrote :

Yes, it's valid.

C99 6.2.1p7:

Each enumeration constant has scope that begins just after the
appearance of its defining enumerator in an enumerator list.

Note that an "enumerator" includes the "= constant" if present, so
"enum { x = x }" is illegal (unless another x was previously declared
in an enclosing scope, but that's ugly so don't do it).

Thanks, that's exactly the reference & explanation that I was seeking.

Francois Grieu