C Programming - Input / Output - Discussion

@All.

Here is my coding part. Please refer it.

p++ goes to 'd'
2nd p++ goes to '\'
and p-2 goes to '% ' at the starting of the pointer

So,
printf becomes printf("%d\n",23)
it treats %d= format specifier
and \n = newline character
So that's why it prints 23 (int).

As far as I know, it's just based on pointers concept. Here "p" is a pointer which points to an address.

An address will be stored in the variable p. Let us assume that address to be 1000. in the above program. The value "%d\n" is assigned to "p" which means that value is stored in the address 1000.. when we give p++ twice.

The value stored in "p" will get incremented twice which means p++=1000+1=1001, again p++=1001+1=1002. So the current value(address) stored in p=1002.."p" is now pointing to the address 1002.

But the value which we assigned "%d\n" is stored in the address 1000 inorder to make p point to that address(1000) we give p-2, where p=1002-2=1000.

Now, p will again point to 1000 so the printf(p-2,23); should be seen like this printf("%d\n",23);. Hence it prints the value 23.

Can anyone explain me what exactly happening inside printf statement's braces? Please.

Please explain me clearly.

Nice @Priyanka.

Thank you @Abhinav Singh.

Please can any one explain clearly? I didn't understand this program.

Nice explanation @Priyanka.

You're absolutely right @Priyanka.

What about the char type man? How can it change from character to integer?