String Problem?

[Davy]

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?
Any suggestions will be appreciated!

Best regards,
Davy

[Walter Roberson]

In article <11************ **********@g44g 2000cwa.googleg roups.com>,
Davy <zh*******@gmai l.com> wrote:

I found
char x[4]={"my"};
can be compiled. But
char x[4];
x={"my"};
can not be compiled. Why?

Because that's how it is.

Initializers on a variable declaration have syntaxes available
that are not available in other places.
--
Camera manufacturers have temporarily delayed introduction of
sub-millibarn resolution bio-hyperdimensiona l plasmatic space polyimaging,
but indications are that is still just around the corner.

[sat]

Ok davy, to understand this,
you need some background on the way C / C++ deals with array names and array
pointers.

char* p1 = "hello";
char p2[] = hello;

now.. p1 and p2 can be used in most places interchangeable .. like
1. p1[0];
2. *(p2+2)
etc etc..
but there are few exceptioins for this.
1. p2++
2. p2 - =10;
3. p2 = p1
4. char* ch = "world" ;
p2 = ch;
ARE NOT VALID

the thumb rule is "you cant do any modication operations to the address
pointed by the array name". You will get an LValue required error.


"Davy" <zh*******@gmai l.com> wrote in message
news:11******** **************@ g44g2000cwa.goo glegroups.com.. .

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?
Any suggestions will be appreciated!

Best regards,
Davy

[werasm]

Davy wrote:

Hi all,

I found
char x[4]={"my"};
can be compiled.
Yes, but it is better to do this:

char x[] = "my";

This way just enough space for "my" is provided.
But
char x[4];
x={"my"};
can not be compiled.
Because the type of x in...

char x[4];

....is effectively...

char* const x;

....which means the value of the pointer "x" is constant - may not be
modified, whereas doing this...

x={"my"}; //- actually x = "my"...

....attemps to modify the pointer x, which is why you get a compiler
error.

Suggestions as follow:

char x[100] = { '\0' };
std::ostrstream os( x, sizeof(x) );
os << "Hooh, hah" << std::ends; //don't forget std::ends;

or...

std::string x("my");

or...

char x[4] = { '\0' };
strcpy( x, "my" );

Remember, the contents of the array which exists at the address
location whereto x points, is modifiable. The address itself (or the
value of the ptr) is not.

Regards,

W


Why?
Any suggestions will be appreciated!

Best regards,
Davy

[Dale]

"Davy" <zh*******@gmai l.com> wrote in news:1128661915 .425811.133260
@g44g2000cwa.go oglegroups.com:

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?

Remember. sweetums, 'x' is a pointer when you declare it that way. If that
assignment actually worked, it would be changing the pointer itself, not
the data that the pointer points to.

Be glad it didn't compile, because the resulting program would have either
crashed or produced bizarre results, and then you'd be sitting there all
day trying to figure out why.

[Mark McIntyre]

On 07 Oct 2005 17:53:50 GMT, in comp.lang.c , Dale <da***@gas.oran ge>
wrote:

"Davy" <zh*******@gmai l.com> wrote in news:1128661915 .425811.133260
@g44g2000cwa.g ooglegroups.com :

char x[4]={"my"};
Remember. sweetums, 'x' is a pointer when you declare it that way.

Actually, no, this is terribly terribly wrong. In neither declaration
is 'x' a pointer - its an array[4] of char.
If that assignment actually worked, it would be changing the pointer itself, not
the data that the pointer points to.

Which would be perfectly legal for a pointer. However for an array,
you simply can't do that.
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >

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[Kid Hugh]

You can use the following:
strcpy(x,"my");
instead of
x={"my"};

x is an array pointer not only a variable.

"Davy" <zh*******@gmai l.com> ????
news:11******** **************@ g44g2000cwa.goo glegroups.com.. .

Hi all,

I found
char x[4]={"my"};
can be compiled.

But
char x[4];
x={"my"};
can not be compiled.

Why?
Any suggestions will be appreciated!

Best regards,
Davy

[Dale]

Mark McIntyre <ma**********@s pamcop.net> wrote in
news:u8******** *************** *********@4ax.c om:

On 07 Oct 2005 17:53:50 GMT, in comp.lang.c , Dale <da***@gas.oran ge>
wrote:

"Davy" <zh*******@gmai l.com> wrote in news:1128661915 .425811.133260
@g44g2000cwa. googlegroups.co m:

char x[4]={"my"};

Remember. sweetums, 'x' is a pointer when you declare it that way.

Actually, no, this is terribly terribly wrong. In neither declaration
is 'x' a pointer - its an array[4] of char.

You're the one who's wrong, here, fucknut -- 'x' is a character pointer
when you declare it that way. That's how C stores an array in memory;
i.e., as a pointer to a chunk of RAM.

Run this little program and observe the output:

int main(void)
{

char x[4]={"my"};

printf("%c\n", x);
printf("%c\n", *x);

return 0;
}

The first printf() will display garbage (because it's actually printing
the value of a pointer) but the second will display the letter 'm'
(because it's printing the value that the pointer points to). If 'x'
were not a pointer then this wouldn't even fucking compile because '*x'
would be considered invalid indirection.

Dumbass.

[John Harrison]

Dale wrote:

Mark McIntyre <ma**********@s pamcop.net> wrote in
news:u8******** *************** *********@4ax.c om:

On 07 Oct 2005 17:53:50 GMT, in comp.lang.c , Dale <da***@gas.oran ge>
wrote:

"Davy" <zh*******@gmai l.com> wrote in news:1128661915 .425811.133260
@g44g2000cwa .googlegroups.c om:

char x[4]={"my"};

Remember. sweetums, 'x' is a pointer when you declare it that way.
Actually, no, this is terribly terribly wrong. In neither declaration
is 'x' a pointer - its an array[4] of char.

You're the one who's wrong, here, fucknut -- 'x' is a character pointer

No x is an array
when you declare it that way. That's how C stores an array in memory;
i.e., as a pointer to a chunk of RAM.
That is just confused. Arrays are stored as 'a chunk of RAM', pointers
don't enter the picture.

Run this little program and observe the output:

int main(void)
{

char x[4]={"my"};

printf("%c\n", x);
printf("%c\n", *x);

return 0;
}

The first printf() will display garbage (because it's actually printing
the value of a pointer) but the second will display the letter 'm'
(because it's printing the value that the pointer points to). If 'x'
were not a pointer then this wouldn't even fucking compile because '*x'
would be considered invalid indirection.
You clearly don't know the rule in C and C++ which allows an array to
convert to a pointer in many cirumstances. x can convert to a pointer,
and that is what it is doing in both your examples above, but that does
not mean that x is a pointer.

Try this program out

int main()
{
char x[99];
printf("%u\n", sizeof x);
}

It x really was a pointer then that would print 4 (the size of a
pointer) but it prints 99 (the sizeof the array). That is because sizeof
is on of the exception to the rule that an array can be converted to a
pointer. There are other exceptions as well but they are rare, so I can
see how you might have missed this.

Dumbass.

If you are going to post to a public forum, you have accept the
possiblity that you might be wrong, and that you might be corrected. It
stops everyone else thinking you are a wanker if you can accept those
occaisions with good grace.

john

[Keith Thompson]

Dale <da***@gas.oran ge> writes:

Mark McIntyre <ma**********@s pamcop.net> wrote in
news:u8******** *************** *********@4ax.c om:

On 07 Oct 2005 17:53:50 GMT, in comp.lang.c , Dale <da***@gas.oran ge>
wrote:

"Davy" <zh*******@gmai l.com> wrote in news:1128661915 .425811.133260
@g44g2000cwa .googlegroups.c om:

char x[4]={"my"};

Remember. sweetums, 'x' is a pointer when you declare it that way.
Actually, no, this is terribly terribly wrong. In neither declaration
is 'x' a pointer - its an array[4] of char.

You're the one who's wrong, here, fucknut -- 'x' is a character pointer
when you declare it that way. That's how C stores an array in memory;
i.e., as a pointer to a chunk of RAM.

Dale, let me offer some friendly advice.

First, don't be rude.

Second, don't be simultaneously rude and wrong.

Arrays are not pointers. Pointers are not arrays. An array is not
stored in memory as "a pointer to a chunk of RAM", it's stored in
memory as an array.

The declaration
char x[4];
declares an array object, with or without an initialization. It does
not declare a pointer object.

I think what's confused you is the fact that an array name, or any
expression of an array type, is implicitly converted to a pointer to
the array's first element in most contexts. (The exceptions are the
operand of the unary "&" or "sizeof" operator, and a string literal in
an initializer.)
Run this little program and observe the output:

int main(void)
{

char x[4]={"my"};

printf("%c\n", x);
printf("%c\n", *x);

return 0;
}

The first printf() will display garbage (because it's actually printing n
the value of a pointer) but the second will display the letter 'm'
(because it's printing the value that the pointer points to). If 'x'
were not a pointer then this wouldn't even fucking compile because '*x'
would be considered invalid indirection.
The first printf() invokes undefined behavior. The expression x,
which is of array type, is implicitly converted to a pointer value
of type char*. Passing this as an argument to printf() with a "%c"
format invokes undefined behavior. It happens to display garbage,
but since you're lying to the compiler it could do anything.

In the second printf(), the subexpression x is again converted to
a value of type char*. Dereferencing this pointer value yields the
character value 'm', so you're right about this one.

And of course you're missing the require "#include <stdio.h>, so
*any* call to printf() invokes undefined behavior. On many (most?)
implementations , it happens to work anyway, but you shouldn't count
on that.

The relationship between arrays and pointers is a common source of
confusion among C newbies. You need to read section 6, "Arrays and
Pointers", of the C FAQ (google "C FAQ" to find it). I see that this
thread is cross-posted to comp.lang.c and comp.lang.c++ (which is
rarely a good idea), but this happens to be an area where C and C++
are similar.
Dumbass.

There's a good chance that a lot of the regulars in these newsgroups
have already killfiled you, or will do so as soon as they see your
article. You may already have permanently damaged your ability to
participate here and to benefit from the advice of experts.

Be embarrassed.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.