C Programming - Pointers - Discussion

Can we declare print statement like printf(str-2, 300);?

MY question is where is "%c" or " "%s"in printf statement. If it is true then how it works?

Nice explanation.

@Nejat.

It's just an assumption. The memory location is allocated randomly by the CPU. It can be any number.

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

Thank you to all.

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

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.

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?

char*str; --> a pointer char points to anywhere, Ex. 100.

str = "%d\n"; --> now points to "%d\n" address of char array. Here there is a address of first element "%d\n".

Let's say that a char length is 2 byte.

So :

% --> address --> 100.
D --> address --> 102.
\ --> address --> 104.
N --> address --> 106.

str++; --> for be increased length of char (2 byte) up now our char array starts 102.

str++; --> for be increased length of char (2 byte) up now our char array starts 104.

printf (str-2, 300) ; for now str is decreased 2 times (2 byte + 2 byte) and it was 100 again. So it points to our char array again.

Return 0;

Please some explain it step by step?