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Memory allocation for an initialized character pointer;

Hi all,

If I have a piece of code something like this

void main(void)
{
char * p1="abcdefghijk lmn";
............... ............... ...............

}

For p1 and whatever it points to , where is the memory allocated. Is
it on heap or stack or
some other common global area. I am assuming it should be either on
stack or some global area. Unless
a user does a malloc storage space cannot be allocated on heap is my
assumption. Can someone
throw some light on how compilers do this.

What if the above piece of code is changed to something like

void main(void)
{
char *p1;
char a[20]="abcdefgh";

strcpy(p1,a);
}

Now where is the space allocated for p1 and whatever it points to
after strcpy.

Thanks in advance.

Regards,
Ar
Sep 30 '08
50 3541
Old Wolf wrote:
On Oct 1, 10:35 am, jameskuy...@ver izon.net wrote:
>String literals are treated differently when they initialize char*
pointers than they are when they initialize char arrays. In the first
case, static memory is allocated for the contents of the string
literal, and the pointer is initialized with a value that points to
the first character of the array. It is not safe to attempt to write
to that array.

In the second case, the contents of the string literal are used to
directly initialize the declared array; no separate storage is
allocated.

Are you sure? My understanding is that the
string literal is still allocated, regardless
of what it's subsequently used for.

(Of course there is no way of telling because
of the as-if rule, it's just a language lawyer
question..)
I thought this was explicitly stated in the standard, but you're right;
the wording about string literals implies that they always result in the
creation of a separate static array during translation phase 7, even if
they are also used to initialize a user-defined array. However, you're
also right in that there's no way for a strictly conforming program to
verify whether or not this has actually occurred; therefore, in practice
this is covered by the as-if rule. If you can't tell whether or not it
exists, a conforming implementation isn't required to provide it.
Oct 5 '08 #51

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