C Programming - Input / Output - Discussion

Hey I don't understand any of your explanations. I guess you guys are complicating it.

As far as I know its 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 adress(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.

Correct me if I'm wrong.

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.

Hi, I think we have to know the pointers first.

In the above program p points to the string "%d\n" and so when the pointer is incremented twice so the pointer is pointed to null position(means it moves to the right of "%d\n").

So in the printf statement the pointer is again moved to the beginning of the string "%d\n"so that the printf statement is similar to printf("%d\n",23);

So the output will be 23.

If any one can explain more detail please do it. Thanks.

@Priyanka.

Your overall answer is correct but there is an error in your explanation, if 1000 is the base address to which p is pointing to i.e., "%" here then p++ will point to "d", and not to the different array, but here for the purpose we need to do p-2,

Consider this program that shows the pointer arithmetic:-

#include<stdio.h>

int main()
{
char *p;
p="%d\n";
p++;
printf("%c",*p);
return 0;
}

Output:- d.

As per my understanding

p="%d\n";
if u print p display one address assume 1000
print *p display =%
p++
address increment 1001x
print *p display =d
p++
address increment 1002
print *p display =\n
after that we print p-2 == means display 1000
but here we write printf(p-2,23) == 23
because printf display last value in argument list
if we write like that print(p-2,23,'C') then display 'C'

@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).

@Paresh is wrong

#include<stdio.h>

int main()
{
char *p;
p="%d\n";
p++;
p++;
int i=1;
int j=3;
printf("%d\n", i,j);
return 0;
}
OUTPUT:
1

We must assign some address value to p, but here we are assigning a string. Can anyone explain why it won't report an error and display 23.

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

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