Every time the function f() is called, it allocates an entirely new set of
memory location, copies "goodbye" in it and returns its base address.
So even if you assign 'A' to its first location, the first character of the
array allocated by f() will still be 'g' if you call it the next time
Thats wh
In all it calls *f() funtion twice once while executing *f()='A' and other
when executing the printf statement.
*f() is returing character pointer so if you print with %c, it prints g and
if you print with %s and do not dereference the function you will be able to
print goodbye.
For reference: ht
it is returning the address of last operation done in the accumulator.
as the last operation is copying to s to it is returning s
you will understand try running this,,...
#include
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}
main()
{
*f()='A';
ya i forgot that...considering that plz explain o/p
i.e
#include
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
return s;
}
main()
{
*f()='A';
printf("%c",*f());
}
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did you forget to return any value from function f() ?
Anurag Sharma
On Mon, Jun 14, 2010 at 7:19 PM, sharad wrote:
> #include
> char *f()
> {char *s=malloc(8);
>strcpy(s,"goodbye");
> }
>
> main()
> {
> *f()='A';
> printf("%c",*f()); }
>
>
> find
#include
char *f()
{char *s=malloc(8);
strcpy(s,"goodbye");
}
main()
{
*f()='A';
printf("%c",*f()); }
find o/p n explain it
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