char *s = "Google"

[Sri Ragaventhirar]

char *s = "Google"

Where this "Google" will be stored?
Why it can't be changed?
like s[1]='h'; -this is giving error
please explain this.

[Marc Boyer]

Le 13-12-2006, Sri Ragaventhirar <sr************ ***@gmail.coma écrit*:

char *s = "Google"

Where this "Google" will be stored?

Somewhere you are not allowed to write in.

Why it can't be changed?

No.

like s[1]='h'; -this is giving error
please explain this.

You are trying to write in a forbiden place.
What you can do is:
char s[]= "Google";
and then, you are allowed to do
s[1]= 'h';

Marc Boyer

[jaysome]

On 13 Dec 2006 00:55:28 -0800, "Sri Ragaventhirar"
<sr************ ***@gmail.comwr ote:

>char *s = "Google"

Where this "Google" will be stored?
Why it can't be changed?
like s[1]='h'; -this is giving error
please explain this.

The answer is in FAQ 1.32.

http://c-faq.com/decl/strlitinit.html

--
jay

[Sri Ragaventhirar]

where this string "google"wil l be stored?
why it can't be changed?

[Sri Ragaventhirar]

where this string "google"wil l be stored?
why it can't be changed?

[Marc Boyer]

Le 13-12-2006, Sri Ragaventhirar <sr************ ***@gmail.coma écrit*:

where this string "google"wil l be stored?

It depends on your platform, ie which compiler, which options,
which OS...

why it can't be changed?

Because the standard states that...
Why the standard did this choice ? I assume this is to allow
this kind of string to be stored in read-only memory, which
could be cheaper in some platform...

Marc Boyer

[Sri Ragaventhirar]

thank you for your info
--------------------------------
Anywhere else, it turns into an unnamed, static array of characters,
and this unnamed array may be stored in read-only memory, and which
therefore cannot necessarily be modified. In an expression context, the
array is converted at once to a pointer, as usual (see section 6), so
the second declaration initializes p to point to the unnamed array's
first element.
-------------------------------
can you explain more about this read only memory(why it is like that
and where it is actually stored)(code segment data segment, stactk
segment???)

[Ian Collins]

Sri Ragaventhirar wrote:

thank you for your info
--------------------------------
Anywhere else, it turns into an unnamed, static array of characters,
and this unnamed array may be stored in read-only memory, and which
therefore cannot necessarily be modified. In an expression context, the
array is converted at once to a pointer, as usual (see section 6), so
the second declaration initializes p to point to the unnamed array's
first element.
-------------------------------
can you explain more about this read only memory(why it is like that
and where it is actually stored)(code segment data segment, stactk
segment???)

It's like that because it is a string literal. It should require const
char*, but const is broken due to backwards compatibility in C.

Where its is stored is up to the implementation.

--
Ian Collins.

[Chris Dollin]

Sri Ragaventhirar wrote:

(several times. DON'T DO THAT. We are not a chatgroup.)

char *s = "Google"

Where this "Google" will be stored?

Somewhere that lasts from when the program starts to when it
ends. (Maybe even longer, but you can't tell.)

Why it can't be changed?

You don't know if it can or can't be. All you know is that
trying to change it gets undefined behaviour: in other words,
the implementation can do whatever it likes.

like s[1]='h'; -this is giving error
please explain this.

What kind of error? A compile-time error or a run-time one?

The compiler may choose to store the string in a region of
read-only memory. That memory may cause an signal to be raised
if you try and change it. BOOM.

The compiler may choose to store the string in a region of
read-only memory. That memory may ignore any attempt to change
it.

The compiler may choose to store the string in a region of
writeable memory. Assigning to elements of the string will
update it.

The compiler may choose to store the string in a region of
read-only memory. Writing to that memory may cause a
hedgehog to come and sign in your ear. Badly.

All four of these are legitimate implementations - even the
fourth, although the shortage of singing hedgehogs makes
it less likely you'll encounter it.

So don't assign into string literals.

--
Chris "We are BROKE-en" Dollin
"We did not have time to find out everything we wanted to know."
- James Blish, /A Clash of Cymbals/

[Chris Dollin]

Sri Ragaventhirar wrote:

thank you for your info
--------------------------------
Anywhere else, it turns into an unnamed, static array of characters,
and this unnamed array may be stored in read-only memory, and which
therefore cannot necessarily be modified. In an expression context, the
array is converted at once to a pointer, as usual (see section 6), so
the second declaration initializes p to point to the unnamed array's
first element.
-------------------------------
can you explain more about this read only memory(why it is like that
and where it is actually stored)(code segment data segment, stactk
segment???)

It's stored somewhere the implementation decides, wherever it's convenient.
Generally speaking one shouldn't need to care. If one /does/ care, one
consults one's local documentation or implementation-specific newsgroup.

Why do you care?

--
Chris "Perikles triumphant" Dollin
"Life is full of mysteries. Consider this one of them." Sinclair, /Babylon 5/