Assigning values to char arrays

[emyl]

Hi all,

here's an elementary question. Assume I have declared two variables,

char *a, **b;

I can then give a value to a like

a="hello world";

The question is, how should I assign values to b? A simple

b[0]="string";

results in a segmentation fault.

Answers greatly appreciated.

Regards, Emyl.

[Richard]

emyl <kw****@yahoo.c omwrites:

Hi all,

here's an elementary question. Assume I have declared two variables,

char *a, **b;

b is a pointer to a char pointer.

>
I can then give a value to a like

a="hello world";

The question is, how should I assign values to b? A simple

b[0]="string";

*b = "s";

>
results in a segmentation fault.

Answers greatly appreciated.

Regards, Emyl.

[Richard Heathfield]

emyl said:

Hi all,

here's an elementary question. Assume I have declared two variables,

char *a, **b;

I can then give a value to a like

a="hello world";

The question is, how should I assign values to b? A simple

b[0]="string";

results in a segmentation fault.

Answers greatly appreciated.

Your definition:

char *a, **b;

reserves sufficient storage for a pointer-to-char named a, and a
pointer-to-pointer-to-char named b.

a="hello world";

assigns a value to this pointer-to-char, the value in question being the
address of the first character in the given string literal.

But b[0]="string"; is a problem, not because there's anything wrong with
the syntax, but because you've made an incorrect assumption.

b is a pointer-to-pointer-to-char, but you haven't pointed it to any
pointers-to-char, so it is currently indeterminate. b[0]="string"; is
*not* an attempt to give a value to b. It is an attempt to give a value to
b[0]. But b[0] is meaningless unless b has a meaningful value.

You can give b a meaningful value in any of several ways, but the most
obvious is to allocate some fresh memory for it:

#include <stdlib.h>

/* allocate memory for some pointers-to-char */
char **cpalloc(size_ t n)
{
char **ptr = malloc(n * sizeof *ptr);
if(ptr != NULL)
{
while(n--)
{
ptr[n] = NULL;
}
}
return ptr;
}

#include <stdio.h>

int main(void)
{
char *a, **b;
a = "what has it got in its pocketses?";
b = cpalloc(2);
if(b != NULL)
{
b[0] = "string";
b[1] = "nothing";

printf("%s\n", a);
printf("%s or %s\n", b[0], b[1]);

free(b);
}
return 0;
}

Be careful. The cpalloc function written above does not allocate storage
for strings, only for a collection of pointers to char. A pointer to char
is sufficient for pointing at a string, but not for storing it.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[CBFalconer]

emyl wrote:

>
here's an elementary question. Assume I have declared two variables,
char *a, **b;
I can then give a value to a like
a="hello world";
The question is, how should I assign values to b? A simple
b[0]="string";
results in a segmentation fault.

"char **b;" declares a pointer to a pointer to char. You could
initialize it with "b = &a;" (provided the a declaration is
present). Then **b is a[0].

However note that your initialization of <a = "hello world";>
leaves a pointing to an unmodifiable string.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[Richard]

Richard <rg****@gmail.c omwrites:

emyl <kw****@yahoo.c omwrites:

>Hi all,

here's an elementary question. Assume I have declared two variables,

char *a, **b;

b is a pointer to a char pointer.

>>
I can then give a value to a like

a="hello world";

The question is, how should I assign values to b? A simple

b[0]="string";

*b = "s";

*slaps cold water on face*

Sorry. That was bullshit. I thought I had included the initialisation
line. See other replies.

Scary.

>>
results in a segmentation fault.

Answers greatly appreciated.

Regards, Emyl.

[emyl]

BIG sigh of relief... Thanks all for your replies. More
specifically,

Richard: Your suggestion is one of the bazillion things I had tried
more or less a
t random. Like you noticed, it won't work. Thanks for taking the
time to answer.

Richard H.: My heartfelt thanks for a crystal clear answer that works
like a charm a
nd is general enough to use it in the broader context of my project.
If I could I'd buy
a beer to you and candy to your kids........... ...

Chuck F.: Thanks a lot for your answer. Of all the simple
possibilities it was the
only one I didn't try, and I guess the only correct one. If I had
thought of that
I'd have had enough of a clue to solve the problem.

[somenath]

On Nov 2, 5:32 am, Richard Heathfield <[e-mail].i nvalidwrote:

emyl said:

Hi all,

here's an elementary question. Assume I have declared two variables,

char *a, **b;

I can then give a value to a like

a="hello world";

The question is, how should I assign values to b? A simple

b[0]="string";

results in a segmentation fault.

Answers greatly appreciated.

Your definition:

char *a, **b;

reserves sufficient storage for a pointer-to-char named a, and a
pointer-to-pointer-to-char named b.

a="hello world";

assigns a value to this pointer-to-char, the value in question being the
address of the first character in the given string literal.

But b[0]="string"; is a problem, not because there's anything wrong with
the syntax, but because you've made an incorrect assumption.

b is a pointer-to-pointer-to-char, but you haven't pointed it to any
pointers-to-char, so it is currently indeterminate. b[0]="string"; is
*not* an attempt to give a value to b. It is an attempt to give a value to
b[0]. But b[0] is meaningless unless b has a meaningful value.

You can give b a meaningful value in any of several ways, but the most
obvious is to allocate some fresh memory for it:

#include <stdlib.h>

/* allocate memory for some pointers-to-char */
char **cpalloc(size_ t n)
{
char **ptr = malloc(n * sizeof *ptr);

if(ptr != NULL)
{
while(n--)
{
ptr[n] = NULL;
}
}

I have one question . Can memset be used as
memset(ptr,0,n) ;
Instead of the while loop ?

return ptr;

[Richard Heathfield]

somenath said:

On Nov 2, 5:32 am, Richard Heathfield <[e-mail].i nvalidwrote:

<snip>

> char **ptr = malloc(n * sizeof *ptr);

> if(ptr != NULL)
{
while(n--)
{
ptr[n] = NULL;
}
}

I have one question . Can memset be used as
memset(ptr,0,n) ;
Instead of the while loop ?

Not unless you can guarantee that the representation of null pointers on
all target platforms is all-bits-zero. I don't recall that the OP
mentioned any platforms. The code I supplied was portable to any hosted
implementation.

In situations where you /can/ use memset, don't bother - just calloc it
instead.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[somenath]

On Nov 2, 11:07 am, Richard Heathfield <[e-mail].i nvalidwrote:

somenath said:

On Nov 2, 5:32 am, Richard Heathfield <[e-mail].i nvalidwrote:

<snip>

char **ptr = malloc(n * sizeof *ptr);

if(ptr != NULL)
{
while(n--)
{
ptr[n] = NULL;
}
}

I have one question . Can memset be used as
memset(ptr,0,n) ;
Instead of the while loop ?

Not unless you can guarantee that the representation of null pointers on
all target platforms is all-bits-zero. I don't recall that the OP
mentioned any platforms. The code I supplied was portable to any hosted
implementation.

In situations where you /can/ use memset, don't bother - just calloc it
instead.

Many thanks for the response.

But my understanding was in pointer context 0 and NULL is converted to
null pointer. And converting to null pointer is compiler
responsibility. So I thought 0 in memset will be converted to null
pointer (which is system specific).

I would request you to correct me as I am feeling I may be
misunderstood some concept.

[santosh]

Ark Khasin wrote:

Richard wrote:

<snip>

I am not to argue who of us two is more of a newbie, but your post
sheds no light on the question asked. Ego bubbling?

This is most hilarious sentence I've read in c.l.c. this year.