Time cost of Haskell `seq` operator

Question

This FAQ says that

The seq operator is

seq :: a -> b -> b

x seq y will evaluate x, enough to check that it is not bottom, then discard the result and evaluate y. This might not seem useful, but it means that x is guaranteed to be evaluated before y is considered.

That's awfully nice of Haskell, but does it mean that in

x `seq` f x

the cost of evaluating x will be paid twice ("discard the result")?

Answer 1

The seq function will discard the value of x, but since the value has been evaluated, all references to x are "updated" to no longer point to the unevaluated version of x, but to instead point to the evaluated version. So, even though seq evaluates and discards x, the value has been evaluated for other users of x as well, leading to no repeated evaluations.