Parse bracketed box expressions

[ftxqxd]

This lets the parser parse expressions like box (1i + 2) as box() (1i + 2). Expressions are parsed greedily in if, while, and for expressions, so that if box(foo) {} {} is parsed as if (box(foo) {}) {}. This is the same behaviour as before this change.

Closes #15386.

[lilyball]

This seems like a bad idea. It's just going to be confusing. The rule implemented here is if box (expr) is followed by something that looks like it could be an expression, treat (expr) as a placement, otherwise treat it as the value to box. As you already stated, this means that if box (foo) { ... } considers the { ... } to be the expression to be boxed. Yes that's the current behavior, but the current behavior is unambiguous: any parenthesized expression after box is always placement.

At the very least, if box (foo) { ... } absolutely must be non-greedy and treat the { ... } as the block for the if expression. But more generally, I think it's confusing to change how we treat the (parenthesized) expression based on whether there's a subsequent expression.

[alexcrichton]

Closing due to inactivity, I also think that changes to the grammar such as this may require an RFC to gain feedback and such.