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Is this type of char array initization legal?

 
 
DomoChan@gmail.com
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      10-24-2008
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
 
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Gennaro Prota
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      10-24-2008
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> 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!


Practical advice: change it to

unsigned char myString[ 200 ] = {} ;

What to you mean by "alleviate" the crash?

PS: the language lawyering behind the first initialization is a
bit more complicated than I have time to explain; consider that
the array elements are of type unsigned char (not char) and that
there was a C standard clarification in this area which C++
hasn't picked up (though your C++ compiler might de-facto
implement it).

<http://groups.google.com/group/comp.lang.c++/msg/d5ac14fbe55de25b>

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Breeze C++ (preview): <https://sourceforge.net/projects/breeze/>
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Salt_Peter
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      10-24-2008
On Oct 24, 12:22*pm, (E-Mail Removed) wrote:
> 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?


Sounds like you see the results of a program in debug mode. The string
literal "" is actually {'\0'} so everything after myString[0] has an
undeterminate value. Nobody cares whether some compiler initializes or
not the rest of the array with some magic value, there is no
requirement that says it must.

>
> 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


An array has a special initializer that looks like so:
unsigned char myString[200] = {'a'};
which initializes all elements with 'a' in this case.
Note the element type of the array (unsigned char) and the type in the
array init list.
Now having said that, the above is an array, not a terminated sequence
of unsigned characters.

Another reason to prefer modern containers:

#include <vector>

// all elements initialized: guarenteed
std::vector< unsigned char > vuc(200);

// all 'a'
std::vector< unsigned char > vuca(200, 'a');
 
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Default User
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      10-24-2008
Pete Becker wrote:

> On 2008-10-24 12:49:25 -0400, Victor Bazarov
> <(E-Mail Removed)> said:


> > 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(myString[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.




Brian

 
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Default User
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      10-24-2008
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
 
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James Kanze
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      10-24-2008
On Oct 24, 10:46*pm, Pete Becker <(E-Mail Removed)>
wrote:
> On 2008-10-24 16:18:56 -0400, Victor Bazarov
> <(E-Mail Removed)> said:
> > 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.)

--
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      10-24-2008
Pete Becker wrote:

> On 2008-10-24 16:29:45 -0400, "Default User"
> <(E-Mail Removed)> said:


> > 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
 
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Default User
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      10-24-2008
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
 
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      10-24-2008
Pete Becker wrote:

> On 2008-10-24 17:53:48 -0400, "Default User"
> <(E-Mail Removed)> said:
>
> > 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 "initializers". 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
 
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James Kanze
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      10-25-2008
On Oct 24, 11:46*pm, Pete Becker <(E-Mail Removed)> wrote:
> On 2008-10-24 17:17:10 -0400, "Default User" <(E-Mail Removed)> said:
> > Pete Becker wrote:


> >> On 2008-10-24 16:29:45 -0400, "Default User"
> >> <(E-Mail Removed)> said:


> >>> 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.


In this case, the words do say what the example shows. But it
has nothing to do with the case at hand, something like:
char str[ 20 ] = "abc" ;
Is this guaranteed to initialize all 20 bytes, or just the first
four?

--
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