Compare Expressions: Fix normalization of exprs by adding zero

[mgrang]

Previously, in order to normalize exprs we added zero only to exprs not
containing any other constants. Also, for binary nodes with + operator we added
+0 and for nodes with * operator we added *1. This resulted in exprs like "p -
i" which do not contain constants but also do not have + or * operator to not
get normalized. So we simply the logic to add zero. Now we simply add a binary
node with the + operator and a 0 child at the root of the AST. This makes the
logic simpler and handles a wider variety of cases.

This fixes issue #933

[sulekhark]

Have we confirmed that constant folding will not fold i + 0 to i ? Will there be some circumstances where we want i + 0 to constant-fold to i, and some other circumstances where we do not want this to happen (ex: to maintain normalization)?

[mgrang]

Have we confirmed that constant folding will not fold i + 0 to i ? Will there be some circumstances where we want i + 0 to constant-fold to i, and some other circumstances where we do not want this to happen (ex: to maintain normalization)?

Yes, I have confirmed that i + 0 will not constant fold to i. For example,

void f(int i) {
  _Nt_array_ptr<char> p : bounds(p, p + 0) = "a";

  if (*(p + 0))
  {}
}

On dumping the preorder AST we see that +0 is preserved:

+
ImplicitCastExpr 0x8927c50 '_Nt_array_ptr<char>' <LValueToRValue>
`-DeclRefExpr 0x8927c08 '_Nt_array_ptr<char>' lvalue Var 0x8927b68 'p' '_Nt_array_ptr<char>'
IntegerLiteral 0x8927f48 'int' 0

For the purpose of the preorder AST normalization we always want to preserve +0.

[sulekhark]

LGTM.

[sulekhark]

LGTM.