As far as I know, the derefence operator * returns the value stored in the pointer address. What I'm confused by is the behavior when the operator is used with pointer of an array. For example,
int a[4][2]; Then a is internally converted to pointer of first element of array of 4 elements of 2 ints. Then which value does *a return? I'm really confused!
The type of a is int[4][2], so the type of *a (or equivalently a[0]) is int[2].
It is not the same as a[0][0]. If you do this:
int a[4][2]; printf("%d\n",*a); The compiler will tell you this:
warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
Since an array (in this case one of type int [2]) is being passed to a function, it decays to a pointer to the first element in this context.
If on the other hand you had **a, that is equivalent to a[0][0] and has type int.
int *, its because expresions of array type in any context other than & or sizeof or related type queries are implicitly converted into pointers. So int[2] becomes int *
&, (b) the operand of sizeof, or (c) a string literal in an initializer used to initialize an array (sub)object. Those are the only exceptions. (The N1570 draft of C11 incorrectly lists a fourth, but the _Alignof operator cannot be applied to an expression.)decltype, typeof, and alignof that exist in various extensions, as well as _Generic. Also possibly offsetof, though in that case conversion to a pointer would have no effect._Generic is converted to a pointer if it's of array type. (gcc agrees; clang doesn't, at least as of version 3.7.1.) offsetof() can't take an array expression as an argument, so it doesn't apply. Extensions might be another matter.This:
int a[4][2]; defined a an array of 4 elements, each of which is an array of 2 int elements. (A 2-dimensional array is nothing more or less than an array of arrays.)
An array expression is, in most contexts, implicitly converted to a pointer to the array object's initial (zeroth) element. (Note the assumption that there is an array object; that has caused some angst, but it's not relevant here.)
The cases where an array expression is not converted to a pointer are:
sizeof;&; and(Compiler-specific extensions like gcc's typeof might create more exceptions.)
So in the expression *a, the subexpression a (which is of type int[4][2]) is implicitly converted to a pointer of type int(*)[2] (pointer to array of 2 ints). Applying unary * dereferences that pointer, giving us an expression of type int[2].
But we're not quite done yet. *a is also an expression of array type, which means that, depending on how it's used, it will probably be converted again to a pointer, this time of type int*.
If we write sizeof *a, the subexpression a is converted from int[4][2] to int(*)[2], but the subexpression *a is not converted from int[2] to int*, so the expression yields the size of the type int[2].
If we write **a, the conversion does occur. *a is of type int[2], which is converted to int*; dereferencing that yields an expression of type int.
Note that despite the fact that we can legally refer to **a, using two pointer dereference operations, there are no pointer objects. a is an array object, consisting entirely of 8 int objects. The implicit conversions yield pointer values.
The implicit array-to-pointer conversion rules are in N1570 section 6.3.2.1 paragraph 3. (That paragraph incorrectly gives _Alignof as a fourth exception, but _Alignof cannot be applied to an expression. The published C11 standard corrected the error.)
Recommended reading: Section 6 of the comp.lang.c FAQ.
*a doesn't have the type int(*)[2]; It has the type int[2].*a from int[2] to int* that doesn't occur. I've updated my answer accordingly -- and you're right about the type of *a.
ais internally converted to pointer of first element of array of 4 elements of 2 ints." is plain wrong!ais called "array" and not "pointer" for a reason? Sorry, but that is a simple logical conclusion (and you gotr the information from tha answers, too). As a programmer, you are required to think for yourself in the first place. Please start with it.