C Programming - Pointers - Discussion

Good explanation, thank you @Adeel.

Thank you @Adeel.

find the output
int s[]='hold';
*s='C';
print f("%s",s);


Can anyone please answer this question?

This program have two error.

int s[]='hold';
*s='C';
print f("%s",s);
error 1:character constent must be one or two character long
error 2:incompatible type conversion

so correct program give below.
char s[]='hold';
*s='C';
print f("%s",s);

The output is = Cold.

Not getting. Please Help me.

1st line of the program makes an array of char pointers(i.e strings in c++), having s[0] = "black" ... etc.
2nd line create in another array having reversed order of strings in array 's'(i.e. the first string in this array is "voilet").
3rd line making **p to point to the same address as **ptr is pointing to.
4th line increases the address of by one to which the **p is pointing to(i.e now **p is pointing to string "pink").
5th line dereferences the next address within the string which was pointing by **p(i.e firstly **p is now pointing to "ink").

And so is the answer.

int main()
{
char *s[]={"black","white","pink","violet"};
char **ptr[] = {s+1, s, s+3, s+2};
char ***p;
p = ptr; p+1;
printf("%c\n", *(*(*++p+1))+3);
return 0;
}


How this above code is printing 'z', can anyone explain this code in detail?

@Alok.

s is an array of character pointer.
s[0] = "black" s[1] = "white" s[2] = "pink" s[3] = "violet"

ptr is an array of pointer to character pointer.
ptr[0] = s+1 ptr[1] = s ptr[2] = s+3 ptr[3] = s+2

p is a pointer to pointer to character pointer.
p = ptr

Now try to understand this
*(*(*++p+1))+3)

first ++p will execute and now p = ptr+1
And *p = ptr[1]

*(*(ptr[1]+1))+3)
*(*(s+1))+3)
*(s[1])+3)
*("white")+3
'w'+3
'z'

Thank you @Adeel.

Then where do we use *** pointer?

Anyone, please explain to me.