C Programming - Pointers - Discussion

Are you want to say in last step that if str-1 it will pt to 102 and again to 100. That means here compiler will take str-2 is for twice time instead of 2 bytes. , right?

Step->1 char*str;

Here str is the char pointer.
Str has own address. Assume i.e. 500.

Step->2 str = "%d\n";

That mean str has at 500 = "%d\n".

Step->3 str++;

Address incremented by 1 because str is char pointer i.e new address will be 501(500+1).

Step->4 str++; Similarly again increment by 1.

Address will be 502 (501+1).

Step->5 printf(str-2, 300);

Here 1st will be evaluate str-2 then print.
Str-2 means decrements two times. Pointer will be at starting location, i.e 500.

Now printf will be like this printf ("%d\n", 300). Then finally print 300.

Step->6 return 0; means program execute successfully.

Here str is char type & how we can assign it to integer type i.e: %d without any conversion.

Thank you to all.

How do we know that the string is first initialized to 300?

Here, Str ++means address increment.

Nice explanation.

Pointers are always incremented by 2 byte. Then how it is possible? can anyone explain this please?

At the starting str will points to the first character of the string which is %.

After executing two times increment operator on str pointer will skip the first two characters of the string which is %d and points to the \.

Inside the printf statement, str-2 make str point to the starting of the string .

So, the whole string will appear in the printf statement.
printf("%d\n",300).

Therefore, the output will be 300.

Nice @Shubham.