What do the * and & operators operate on if the argument is complex?

Question

Simply, is

&someObject->someAttribute.someMember;

equivalent to

&(someObject->someAttribute.someMember);

or

(&someObject)->someAttribute.someMember;

or

(&(someObject->someAttribute)).someMember;

Or should you really put explicit parenthesis there just to be safe?

Answer 1

It is equivalent to:

&((someObject->someAttribute).someMember)

The -> and . have equal precedence and are left-associative. The unary-& has lower precedence.

If you have very complex expressions, then you should certainly use parentheses to group things and clarify your code. This isn't particularly complex, though; it is quite common to see code like this in C and C++.

Answer 2

See Wikipedia

Scroll down to the operators section. Operators in a higher box in the chart are considered before operators in a lower box. Also see the rules for associativity.

Or should you really put explicit parenthesis there just to be safe?

This is somewhat of a preference issue but as a general rule, if you're having any trouble figuring it out, then someone else will too. Put parenthesis when in doubt.

Answer 3

If u want address of the someMember then &(someObject->someAttribute.someMember) OR &someObject->someAttribute.someMember.

First someObject->someAttribute and second to its member somMember and third is & address

struct xxx{ int xx; }; struct temp{ int x; struct xxx ptr; };

    struct temp t1 = {125,{127}};
    struct temp *t2 = &t1;
    printf("%d %d",t2->x,*&t2->ptr.xx);