C Programming - Arrays - Discussion

Please give me a brief description.

Here we will identify the answer depending on **ptr. At first **ptr points to first element i.e '0'. After incrementind ptr++ that address 2nd element. So in the options 'C' option have '1'. So we guess 'c' as the answer.

Please anyone explain this in detail.

Please explain any one.

p[]=(0,1,2,3,4);
ptr is point to pointer operation which will increament index position by one.

So , Now its point to 1. So ptr 1 and p 0

So for first printf statement ptr-p, *ptr-arr, **ptr => 1-0,1-0,1

Same way fellow for next by increment ptr where *ptr and ptr give the value in the index postion.

Please give me solid answer with fully explanation.

I can't understand the last row. Why's it 3 4 4?

++*ptr; this statement changes the p[3], to points to arr[4].

So the last printf prints 3 4 4.

p is an array of pointers. ptr is a pointer to a pointer, initailly referring to p.
Step 1:
ptr++
ptr is incremented by 1, i.e. ptr now points to p+1 or (arr+1)
ptr-p=1, *ptr=arr+1, so *ptr-arr=1, **ptr=*(arr+1)=1
Step 2:
*ptr++
is *(ptr++), i.e. ptr=ptr+1=arr+2, the * part is not of any work here.
ptr-p=2, *ptr=arr+2, so *ptr-arr=2, **ptr=*(arr+2)=2
Step 3:
*++ptr
Similar to step 2, *++ptr=*(++ptr)
ptr-p=3, *ptr=arr+3, so *ptr-arr=3, **ptr=*(arr+3)=3
Step 4:
++*ptr
is ++(*ptr), i.e. ptr now points to arr+4, but point to be noted that *p also changes,since ptr is pointer to p. Here ptr does not change its reference position unlike all the other steps, but changes the value it is pointing to
*p[]={arr, arr+1, arr+2, arr+4, arr+4}
so, ptr-p=3, *ptr=p[3]=arr+4, so *ptr-arr=4, **ptr=*(arr+4)=4
Hope this helps

I also confused man some body help.