CyberPsychotic wrote:
> ~ With
> ~
> ~ extern char lolo[];
> ~
> ~ the linker will treat the address which is associated with the symbol
> ~ `lolo' as the address of the first character in the array.
> ~
> ~ OTOH, with
> ~
> ~ extern char *lolo;
> ~
> ~ the linker will treat the address which is associated with the symbol
> ~ `lolo' as the address of a pointer (i.e. a 4/8 byte integer).
>
> address of a pointer? could you clarify this.
`char *lolo' implies an extra level of indirection. The address which
is stored in the relocation table of the .o file will be equivalent to
`&lolo', while with `char lolo[]' the address would be equivalent to
just `lolo'.
> so in this case, it just would store in variable lolo, the address of a
> pointer, right, while, in case `char lolo[]="foobar";', it would store
> there the address of the first character in array. Hmmm.. but it doesn't
> make the difference to me..
Try compiling the following to assembler:
char lolo1[] = "Hello";
char *lolo2 = "World";
The symbol lolo1 will refer to the address of the `H' in `Hello'. The
symbol lolo2 will refer to the address of 4 bytes (on i386) of data,
which contains the address of the `W' in `World'.
> ~ Also, with `extern char *lolo', you would be able to do
> ~
> ~ lolo = <something>;
> ~
> ~ i.e. assign a different value to the pointer variable `lolo'. However,
> ~ with `extern char lolo[]', you can't do this, as `lolo' is effectively
> ~ a constant and not a variable.
>
> ahmmm.. now it makes sence. So in my case when I had:
>
> char foo[]="....";
>
> it just had a constant, foo, which pointed to a string, which contained
> "....",
Yes.
> and when I used extern char *foo; in another code, compiler
> happily pointed foo to that place of memory, right?
The compiler assumed that foo was the address of a pointer, so it
would compile code to perform an extra level of indirection.
--
Glynn Clements <[EMAIL PROTECTED]>
