skipping a structure member while initializing

[Mik0b0]

Hallo.
Let's say, there is a structure
struct struct10{
int a1;
int a2;
int a3;
int a4;
}count[2]={
{10,20,30,40},
{50,60,70,80}
};

How can I skip a member, e.g. a2 while initializing?
Thanks a lot!

[Chris Torek]

>On Mar 10, 10:23 am, Matt Kowalczyk <matt5...@comca st.netwrote:
[edited to mininal example]

>>struct foo {
int a1;
int a2;
} foobar[10] = {
[0].a1 = 10,
[2] = {20, 2}
};

In article <11************ **********@v33g 2000cwv.googleg roups.com>
Rajesh S R <SR**********@g mail.comwrote:

>I am not able to understand how the above initialisation is related
with Compound literals ...

It is unrelated to "compound literals". The above is an example of
"designated initializers".

A "compound literal" is a syntactic entity, consisting of two
main parts. The first part looks exactly like a cast: it has
a type-name enclosed in parentheses. The second part looks
exactly like an aggregate initializer: it has values enclosed
in braces. So:

(int) { 2 }

or:

(struct foo) { 9, 7 }

are both examples of compound literals.

A "designated initializer" is also a synactic entity, consisting
of two main parts. The second part is any ordinary initializer
(which may or may not be enclosed in braces as usual); the first
part is the designator, consisting of things like "[2] =" or
".a1 =".

You can combine designated initializers with compound literals,
as in:

(struct foo) { .a2 = 42 }

Here the compound literal contains a designated initializer, so
this makes a "struct foo" object -- whose storage duration depends
on where the compound literal appears in the source -- whose value
is { 0, 42 }, i.e., the a1 member is initialized to 0, and the a2
member is initialized to 42.

Both designated initializers and compound literals are C99-specific,
hence not supported by many C (C89 or C95) compilers. (C89 and
C90 are the same language: C89 here refers to the 1989 ANSI C
standard, X3.159-9899, while C90 refers to the 1990 ISO C standard,
ISO 9899-1990. While the two standards differ, it is mainly in
section numbering; both describe the same language.)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[Rajesh S R]

On Mar 10, 10:49 pm, Chris Torek <nos...@torek.n etwrote:

On Mar 10, 10:23 am, Matt Kowalczyk <matt5...@comca st.netwrote:

[edited to mininal example]

>struct foo {
int a1;
int a2;
} foobar[10] = {
[0].a1 = 10,
[2] = {20, 2}
};

In article <1173542926.948 439.307...@v33g 2000cwv.googleg roups.com>
Rajesh S R <SRRajesh1...@g mail.comwrote:

I am not able to understand how the above initialisation is related
with Compound literals ...

It is unrelated to "compound literals". The above is an example of
"designated initializers".

A "compound literal" is a syntactic entity, consisting of two
main parts. The first part looks exactly like a cast: it has
a type-name enclosed in parentheses. The second part looks
exactly like an aggregate initializer: it has values enclosed
in braces. So:

(int) { 2 }

or:

(struct foo) { 9, 7 }

are both examples of compound literals.

A "designated initializer" is also a synactic entity, consisting
of two main parts. The second part is any ordinary initializer
(which may or may not be enclosed in braces as usual); the first
part is the designator, consisting of things like "[2] =" or
".a1 =".

You can combine designated initializers with compound literals,
as in:

(struct foo) { .a2 = 42 }

Here the compound literal contains a designated initializer, so
this makes a "struct foo" object -- whose storage duration depends
on where the compound literal appears in the source -- whose value
is { 0, 42 }, i.e., the a1 member is initialized to 0, and the a2
member is initialized to 42.

Both designated initializers and compound literals are C99-specific,
hence not supported by many C (C89 or C95) compilers. (C89 and
C90 are the same language: C89 here refers to the 1989 ANSI C
standard, X3.159-9899, while C90 refers to the 1990 ISO C standard,
ISO 9899-1990. While the two standards differ, it is mainly in
section numbering; both describe the same language.)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

Thank you very much for your explanations.
I understand it now.

[David Thompson]

On 10 Mar 2007 17:49:49 GMT, Chris Torek <no****@torek.n etwrote:

<snip good but uncharacteristi cally brief answer <G>>

Both designated initializers and compound literals are C99-specific,
hence not supported by many C (C89 or C95) compilers. (C89 and
C90 are the same language: C89 here refers to the 1989 ANSI C
standard, X3.159-9899, while C90 refers to the 1990 ISO C standard,

I doubt X3 (now X3'') or even ANSI will remain in existence that long.

ISO 9899-1990. While the two standards differ, it is mainly in
section numbering; both describe the same language.)

You wanted X3.159- *1989* of course.

And to be picky, it's ISO/IEC 9899:1990, because ISO uses hyphen for
parts and colon for year, like 1539-1:year and 1539-2:year for Fortran
and 13818-MANY:year for MPEG.