C Programming - Pointers - Discussion

Hi everyone!

Please explain, how come char ptr gives an integer as result without typecasting?

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

What will be the output char ch=300;?

As it is unsigned it exceeds the range, what is the answer and why please explain.

Please some explain it step by step?

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;

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?