Question

According to the manual of quote(expr):

expr: any syntactically valid R expression

While quote(x==y) returns a call x==y successfully, quote(x=y) fails:

Error in quote(x = y) : supplied argument name 'x' does not match 'expr'

Both x=y and x==y are syntactically valid R expressions, aren’t they? Why quote() fails on x=y?

Answer 1

As it says in ?"=":

The operator ‘<-’ can be used anywhere, whereas the operator ‘=’ is
only allowed at the top level (e.g., in the complete expression typed
at the command prompt) or as one of the subexpressions in a braced
list of expressions.

Using = in an argument to quote is not using it at the top level, so you have to put it in braces or parentheses, but you still have to be careful how you evaluate this expression, since the rules above will still apply.

quote({x=y})
quote((x=y))

To address a comment:

As Gavin Simpson said: basically, the “top level” is when you type or run the code at the prompt and is not within a function call. Take z = quote(expr=x) for example. z = quote(...) is evaluated at the top level, but expr=x is not because it’s inside a function call.

In quote(expr=x), = is being used to assign the value of x to the function argument expr; so you’re no longer working at the top-level, you’re constructing a function argument list (pairlist). The reason quote(x=y) fails is because quote doesn’t have an x argument.

The top level context is described briefly in R Internals, in Section 1.4, Contexts.