Evaluation order of &a[0]-CodePudding

Assume one dimensional array a={1,2,3,4} with base address as 100.Find the value of p.

P=&a[0];

I know it will give tha base address of 0th element of 1D array. But in which order c compiler will evaluate this in first way or second?

FIRST:

P=&a[0];
P=&*(a 0);
P=&*a;

Or Second:

P=&a[0];
P=*&(a 0);
P=*&a;

CodePudding user response：

Postfix operators bind stronger than prefix operators. So P = &a[0]; is parsed as

P = &(a[0]);

Which can be transformed as

P = &(*(a   0));

Equivalent to

P = &(*a);

Also equivalent to

P = &*a;

If a is a pointer or an array, the expression ultimately boils down to:

P = a;

CodePudding user response：

I am unclear what you are actually asking, because the answer to the question I read is so obvious.
I will just phrase my thoughts on your shown codes.

P=&a[0];, probably the expression you want to discuss.
P=&*(a 0);, probably your first step of interpretation, but inconsequently ()ed.
P=&(*(a 0));, my guess of what you would write if you were more rigorously ()ing. P=&*a;, unclear, does not help in the analysis. I'll ignore it.

Lets look at the inner *(a 0); an expression including a, which probably is an array or a pointer to something. a 0 can be used as a pointer to the same. *(a 0) would dereference that pointer, getting you the first element as operand for further steps. &(...) then gets you the address of that first element.
OK, makes sense.

The alternative would be:

P=*(&(a 0));, with an inner &(a 0).
This is the expression (a 0), which could be used as a pointer (see above) but not as a pointer variable. I.e. it cannot be assigned to itself and it does not have an address which can be expressed with &(...). That is where this path ends.

So one path which makes sense, one which does not, seems clear to me.