When we talk about dereference, is it necessary that * should be used in it? If we access the referent of the pointer in some other way, can it be considered as dereferencing a pointer or not, like:
char *ptr = "abc" ;
printf( "%c" , *ptr ); // Here pointer is dereferenced.
printf( "%s" , ptr ); // What about this one?
That is the first part of my question.
Now if printf( "%s" , ptr ) is an example of dereferencing then kindly answer the following part of my question too.
K&R says
a "pointer to void" is used to hold any type of pointer but cannot be dereferenced itself
Hence,
char a = 'c' ;
char *p = &a ;
void *k = &a;
printf( "\n%c\n" , *p );
printf( "\n%c\n" , *k );
Does not compile, complier gives error
In function ‘main’: warning: dereferencing ‘void *’ pointer error: invalid use of void expression
But if we use
char *a = "c" ;
char *p = a ;
void *k = a;
printf( "\n%c\n" , *p );
printf( "\n%s\n" , k );
It compiles and works. Which means void pointer can be dereferenced - we have got the object pointer is referring to.
If that's the case then what does K&R above mentioned quote means in this context?
Thanks for your time.
No. what you have is "undefined behaviour" - the C language standard does not say what should happen. In your case, it "works", but for another user/compiler/platform it might not. Your statement:
printf( "\n%s\n" , k );
is equivalent to:
int k = 42;
printf( "\n%s\n" , k );
and is equally undefined.