Using pointers with multidimensional Arrays

[DazedNConfused]

I'm having trouble in understanding this.For example,
If we have an array BEANS[j]:

*(*(beans+i)+j)-why does *(beans+i)refer to an address of the element and not it's value?Afterall,'*'is INDIRECTION operator.Why does only the outmost '*' behave like indirection operator?

I have another question.This two expresions are the same:

*(beans+j)-this one I understand
(*(beans+i))[j]-why is indirection operator in parentheses,shouldn't it be just the other way:
*((beans+i)[j])

Thank you for helping me out,so I can finaly skip the confused part in my name

 

[VincentP]

beans is identical to *(beans+i). Thus

beans[j] or *(beans+j) or *(*(beans+i)+j) or (*(beans+i))[j]

are all equivalent.

Hope this helps,

Vincent

 

[VincentP]

Sorry about that. Here it is again:

beans is identical to *(beans+i). Thus

beans[j] or *(beans+j) or *(*(beans+i)+j) or (*(beans+i))[j]

are all equivalent.

Hope this helps,

Vincent

 

[DazedNConfused]

Thank you for your replie,but what I want to know is:

why is *(beans +i) identical to beans.Why doesn't *(beans+i)refer to value instead of an adrress?Why isn't '*' redeferencing operator here?

Again,thank you

 

[mingis]

Again this multidimensionality...
The meaning of operators '*' and '[]' is the same - apply one level of indirection to the operand. Two-dimensional array
int beans[5][10];
has double indirection. The internal '*' or '[]' takes out only one indirection, so beans[3] or *(beans+3) gives a pointer to single-dimensional array of 10 elements, the 3-rd element of beans[5][10] with the type (int *). Elements of the lowest level can be accessed as beans[3][8] or *(*(beans+3)+8) - with double indirection.

Vincent, your turn...