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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T02:06:31+00:00 2026-06-05T02:06:31+00:00

I’m a big fan of the Clojure / functional approach of programming with immutable

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I’m a big fan of the Clojure / functional approach of programming with immutable values.

However I’m unsure if a delay should be considered as an immutable value (assuming that you delay a pure function). I’m particularly interested in the case where there are one or more delays in a larger immutable data structure.

e.g. a vector containing a delay:

[1 2 (delay (reduce + (range 1000)))]

As far as I can see this behaves as if it is an immutable value in the sense that you can’t see the result of the delay until you force its evaluation – and then the result is cached and the value can never change after that.

Are there any issues with treating a delay as an immutable value in this way?

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  1. Editorial Team
    Editorial Team
    2026-06-05T02:06:33+00:00Added an answer on June 5, 2026 at 2:06 am

    A delay models what’s usually called a thunk, a reference to a yet-to-be-evaluated expression that is replaced with its result once forced, and is thereafter immutable. Haskell uses such internally mutable thunks to model non-strict evaluation. The expression [1, 2, foldl1 (+) [0..1000]] is nominally the same as its explicitly delayed equivalent in a language with strict evaluation.

    Provided of course the function used in the delay object is pure, there is no harm in treating it as immutable. You can think of this in a couple of ways:

    • A pure function can, by definition, be replaced with its result.

    • Local mutation (in this case, of the delay object) does not make a function impure.

    Of course, Clojure does not distinguish between pure and impure functions, so it’s up to you as a developer to be diligent about it.

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