The code
int n = 25;
int *p = &n;
printf("%x\n %d\n %x\n", p, p[0], p[1]);
returns:
\<adress-of-p
25
\<adress-of-p>
Of course I would never do this but in K&R states that
"if pa is a pointer, expressions may use it with a subscript; pa[i] is identical to *(pa i).
so I was curious.
CodePudding user response:
This statement
printf("%x\n %d\n %x\n", p, p[0], p[1]);
invokes undefined behavior by two reasons.
The first one is that to output a pointer you should use a correct conversion specifier. The second one is that you may not dereference a pointer like this p[1]
that does not point to a valid object.
Instead you could write for example
printf("%p\n %d\n %p\n", ( void * )p, p[0], ( void * )( p 1 ) );
CodePudding user response:
When you evaluate p[1]
in your code, you are invoking undefined behavior so your program can do anything.
It is undefined behavior because p
points at n
which is just a single integer, not an array of integers. So p[0]
is n
, but p[1]
is undefined. Basically this is an array overflow bug.
CodePudding user response:
Your program has undefined behaviour, because it dereferences a pointer that doesn't point to anything.
The particular symptoms you have experienced are consistent with reading the next variable in the function
CodePudding user response:
The p[n] syntax is syntactic sugar for *(p n).
So p[0] is *(p 0), which is *p, and since p points to n, *p is the value of n, which is 25.
Now, p[1] is *(p 1). So in theory what you would get is the next integer in memory following n. In this particular case, the next thing in memory happened to be p itself, which is the address of n. (Note that this isn't guaranteed, your compiler just chose to arrange things that way.)