evaluation - Why does ++++x return an increment of 2 when the value of x is only incremented by 1?

This line returns 3:

x = 1; ++++x

However, the value of x after the increment is only 2. Similarly, this line returns 5, while the value of x is again only 2.

x = 1; ++++++++x

Why does it return 3 and 5 respectively in the above examples?

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(This question is intended as a puzzle, and to encourage people to think through an opaque evaluation chain.)

Answer

TracePrint will show you what happens:

PreIncrement takes it's argument x, evaluates it (let's call the result result), then evaluates x = result+1. Note that PreIncrement has HoldFirst.

Now ++(++x) evaluates ++x first yielding 2, then evaluates (++x) = 2+1 resulting in an error (trying to assign to PreIncrement) and returning 3.

This also explains why adding yet another layer of PreIncrement will increment the result again.

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Here's a self-implemented ++ to make the above more clear. The behaviour is exactly the same: