Is this type of char array initization legal?

[DomoChan]

the code below will compile in visual c++ 2003, but im not sure its
valid.

unsigned char myString[200] = "";

after this line executes, all the bytes within myString are indeed set
to '0's' but is this really valid c++ or c? where can I find out how
this is implemented?

Im concerned because I had a 3rd party library wrapper which was
crashing, and I was able to alleviate the crash by changing the
initialization method from the above to ...

unsigned char myString[200];

memset( myString, 0, sizeof( myString ) );

any guidance is greatly appreciated!
-Velik

[Pete Becker]

On 2008-10-24 16:29:45 -0400, "Default User" <de***********@ yahoo.comsaid:

Pete Becker wrote:

>On 2008-10-24 12:49:25 -0400, Victor Bazarov
<v.********@co mAcast.netsaid:

>>The only reason it could be crashing is if the compiler wastn't
providing proper initialisation for the array. What you could do
is revert this to what it was and put an assertion to see if it's
indeed the problem:

unsigned char myString[200] = "";
#ifndef NDEBUG
for (int iii = 0; iii < 200; ++iii)
ASSERT(myStri ng[iii] == 0);
#endif

Hmm, is this required if the char array has automatic storage
duration? I have always assumed that it wasn't, that only the
characters corresponding to characters in the initializer would be
initialized, but it doesn't seem completely clear from a quick glance
at the standard.

Interesting. I also didn't find anything explicit regarding string
literals, just the usual "fewer initializer" rule:

If there are fewer initializers in the list than there are
members in the aggregate, then each member not explicitly
initialized shall be value-initialized (8.5).

The C standard does contain such wording (assuming that it didn't
change from the draft standard):

[#21] If there are fewer initializers in a brace-enclosed
list than there are elements or members of an aggregate, or
fewer characters in a string literal used to initialize an
array of known size than there are elements in the array,
the remainder of the aggregate shall be initialized
implicitly the same as objects that have static storage
duration.

C89's wording was similar to the C++ standard, but had further examples
to show that initialization with a string literal is identical to a
brace-enclosed list of the characters, which would amount to the same
thing.

So C99 made it crystal clear, presumably because C90 wasn't. <g>

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Default User]

Victor Bazarov wrote:

Pete Becker wrote:

On 2008-10-24 12:49:25 -0400, Victor Bazarov

Hmm, is this required if the char array has automatic storage
duration? I have always assumed that it wasn't, that only the
characters corresponding to characters in the initializer would be
initialized, but it doesn't seem completely clear from a quick
glance at the standard.

8.5.1/7:
<<If there are fewer initializers in the list than there are members
in the aggregate, then each member not explicitly initialized shall
be value-initialized (8.5).>>

To me it's pretty clear. Each character from the literal initialises
its respective element of the array. The terminating null character
does initialise the corresponding element of the array too. If there
are fewer characters in the literal than elements in the array (which
is an aggregate), the rest of the array elements are zero-initialised.

I'm sure it was intended, and it's always been that way in C, but
there's some haziness. A string literal isn't a list, exactly. As I
pointed out in another reply, the C99 standard added wording to make
that explicit, and the older standard had wording that confirmed that a
string literal was the same as a character list for intialization
purposes.

In my experience, a C++ compiler that didn't zero out the rest of the
array would be unusual. The OP can certainly check to see if that's the
problem.

Brian

[James Kanze]

On Oct 24, 10:46*pm, Pete Becker <p...@versatile coding.com>
wrote:

On 2008-10-24 16:18:56 -0400, Victor Bazarov
<v.Abaza...@com Acast.netsaid:

8.5.1/7:
<<If there are fewer initializers in the list than there are members in
the aggregate, then each member not
explicitly initialized shall be value-initialized (8.5).>>

Well, yes, that's the requirement for aggregate
initialization. But does aggregate initialization apply here?
Aggregate initialization is indicated by "a brace-enclosed,
comma-separated list ...", which isn't present here. I don't
see anything in 8.5.2 [dcl.init.string] that says that
aggregate initialization is used, although the "optionally
enclosed in braces" hints that it may be.

Well, I rather think it was intended, although you do have a
point. Probably a defect in the standard. (Maybe I should
raise an issue. I'd love to say that it was just editorial, and
just push it off on you, but I think it's a little too much for
that.)

--
James Kanze (GABI Software) email:ja******* **@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[Default User]

Pete Becker wrote:

On 2008-10-24 16:29:45 -0400, "Default User"
<de***********@ yahoo.comsaid:

C89's wording was similar to the C++ standard, but had further
examples to show that initialization with a string literal is
identical to a brace-enclosed list of the characters, which would
amount to the same thing.

So C99 made it crystal clear, presumably because C90 wasn't. <g>

It was, I think, clear enough. It just wasn't all put together nicely.
They had examples to show that:

char str[] = "XYZ";

Was equivalent to:

char str[] = {'X', 'Y', 'Z', '\0'};

So if a string literal is equivalent to a brace-enclosed list of
characters, then the "fewer initializer" rule would come into play to
finish out an array larger than the string literal with '\0'.


Brian

[Pete Becker]

On 2008-10-24 17:17:10 -0400, "Default User" <de***********@ yahoo.comsaid:

Pete Becker wrote:

>On 2008-10-24 16:29:45 -0400, "Default User"
<de*********** @yahoo.comsaid:

>>C89's wording was similar to the C++ standard, but had further
examples to show that initialization with a string literal is
identical to a brace-enclosed list of the characters, which would
amount to the same thing.

So C99 made it crystal clear, presumably because C90 wasn't. <g>

It was, I think, clear enough. It just wasn't all put together nicely.
They had examples to show that:

char str[] = "XYZ";

Was equivalent to:

char str[] = {'X', 'Y', 'Z', '\0'};

So if a string literal is equivalent to a brace-enclosed list of
characters, then the "fewer initializer" rule would come into play to
finish out an array larger than the string literal with '\0'.

Examples in standards are not normative. That is, they do not impose
requirements. They illustrate what the words say. If the words don't
say it, examples don't make it true.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Pete Becker]

On 2008-10-24 17:11:10 -0400, James Kanze <ja*********@gm ail.comsaid:

On Oct 24, 10:46Â*pm, Pete Becker <p...@versatile coding.com>
wrote:

>On 2008-10-24 16:18:56 -0400, Victor Bazarov
<v.Abaza...@co mAcast.netsaid:

>>8.5.1/7:
<<If there are fewer initializers in the list than there are members in
the aggregate, then each member not
explicitly initialized shall be value-initialized (8.5).>>

>Well, yes, that's the requirement for aggregate
initialization . But does aggregate initialization apply here?
Aggregate initialization is indicated by "a brace-enclosed,
comma-separated list ...", which isn't present here. I don't
see anything in 8.5.2 [dcl.init.string] that says that
aggregate initialization is used, although the "optionally
enclosed in braces" hints that it may be.

Well, I rather think it was intended, although you do have a
point. Probably a defect in the standard. (Maybe I should
raise an issue. I'd love to say that it was just editorial, and
just push it off on you, but I think it's a little too much for
that.)

Me, too. <g>

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Default User]

Pete Becker wrote:

Examples in standards are not normative. That is, they do not impose
requirements. They illustrate what the words say. If the words don't
say it, examples don't make it true.

The question is whether the words do say it or not. The examples
certainly help clarify the situation. On the other hand, adding
explicit wording to the next standard indicates that not everyone was
so convinced.

Brian

[Pete Becker]

On 2008-10-24 17:53:48 -0400, "Default User" <de***********@ yahoo.comsaid:

Pete Becker wrote:

>Examples in standards are not normative. That is, they do not impose
requirements . They illustrate what the words say. If the words don't
say it, examples don't make it true.

The question is whether the words do say it or not. The examples
certainly help clarify the situation.

No. Pretend the examples aren't there. What do the words say? If they
don't impose some particular requirement, then it's not a requirement,
even if examples imply that it is.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Default User]

Pete Becker wrote:

On 2008-10-24 17:53:48 -0400, "Default User"
<de***********@ yahoo.comsaid:

Pete Becker wrote:

Examples in standards are not normative. That is, they do not
impose requirements. They illustrate what the words say. If the
words don't say it, examples don't make it true.

The question is whether the words do say it or not. The examples
certainly help clarify the situation.

No. Pretend the examples aren't there. What do the words say? If they
don't impose some particular requirement, then it's not a
requirement, even if examples imply that it is.

The question is, does the requirement for filling out an array when
there is fewer elements in the initializer list than there are in the
declarared array apply when the initializer is a string literal? It
doesn't specifically say that it does, but is it a legitimate
interpretation?

Searching back in comp.std.c, I find that apparently this issue was
submitted as a Defect Report / Technical Corrigendum that "clarifies
that a string literal is equivalent to a brace-enclosed
list of characters."

So the issue has come up before there, and that report is probably why
C99 explicitly said so.

It should be noted that the C++ standard does refer to the individual
characters of a string literal as "initialize rs". Whether that's
sufficient or not is hard to say.

I don't have time right now to search comp.std.c++ (I'm off to the
hockey game).

Brian

[blargg]

In article
<79************ *************** *******@y29g200 0hsf.googlegrou ps.com>, James
Kanze <ja*********@gm ail.comwrote:

On Oct 24, 10:46=A0pm, Pete Becker <p...@versatile coding.com>
wrote:

On 2008-10-24 16:18:56 -0400, Victor Bazarov
<v.Abaza...@com Acast.netsaid:

8.5.1/7:
<<If there are fewer initializers in the list than there are members in
the aggregate, then each member not
explicitly initialized shall be value-initialized (8.5).>>

Well, yes, that's the requirement for aggregate
initialization. But does aggregate initialization apply here?
Aggregate initialization is indicated by "a brace-enclosed,
comma-separated list ...", which isn't present here. I don't
see anything in 8.5.2 [dcl.init.string] that says that
aggregate initialization is used, although the "optionally
enclosed in braces" hints that it may be.

Well, I rather think it was intended, although you do have a
point. Probably a defect in the standard. (Maybe I should
raise an issue. I'd love to say that it was just editorial, and
just push it off on you, but I think it's a little too much for
that.)

If it's not required, then the following would result in a[2] and a[3]
having const indeterminate values, a rather odd situation:

char const a [4] = "X";