How does C interpret int in pointers and why it differ from char pointer? -

#include<stdio.h> void main() {   int *c=12345;   printf("dec:%d  hex:%x", c ,&c); // (1)  } 

the line (1) prints

dec:12345  hex:8af123 

where 8af123 address random , machine dependent.

when put in

printf("dec:%d", *c); // (1) 

it fails, obviously.

so question per theoretical concept:

• *c should holds value 12345 not. why?

and in code:

#include<stdio.h>   void main()   {     char *c='a';     printf("address store in c:%d  value point c:%c address of c:%x", c,*c ,&c); //focus line   }   

output is:

adderess store in c:9105699 value point c:a address of c:8af564 

• why storing 'a' in *c instead in c?

i using gcc compiler 4.4.3.

int *c=12345; 

you made pointer points (probably) invalid address 12345.

if pass c (no *) printf, you're passing pointer itself. therefore, printf sees number 12345, , prints that. has no way of knowing that's supposed pointer.

if pass *c, printf, you're dereferencing pointer – you're passing value @ memory address 12345.
since that's not valid address, dereference operation crash (to precise, behave undefinededly) before ever gets printf.