C Interview Questions and Answers


What is wrong with this declaration?

What's wrong with this declaration?

char* p1, p2;

I get errors when I try to use p2.

Nothing is wrong with the declaration--except that it doesn't do what you probably
want. The * in a pointer declaration is not part of the base type; it is part of
the declarator containing the name being declared That is, in C, the syntax and
interpretation of a declaration is not really

type identifier ;

but rather

base_type thing_that_gives_base_type ; where ``thing_that_gives_base_type''--the
declarator--is either a simple identifier, or a notation like *p or a[10] or f()
indicating that the variable being declared is a pointer to, array of, or function
returning that base_type. (Of course, more complicated declarators are possible
as well.)

In the declaration as written in the question, no matter what the whitespace suggests,
the base type is char and the first declarator is ``* p1'', and since
the declarator contains a *, it declares p1 as a pointer-to-char. The declarator
for p2, however, contains nothing but p2, so p2 is declared as a plain char, probably
not what was intended. To declare two pointers within the same declaration, use

char *p1, *p2;

Posted by:Richards