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498 points azhenley | 1 comments | | HN request time: 0s | source
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EastLondonCoder ◴[] No.45770007[source]
After a 2 year Clojure stint I find it very hard to explain the clarity that comes with immutability for programmers used to trigger effects with a mutation.

I think it may be one of those things you have to see in order to understand.

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rendaw ◴[] No.45770924[source]
I think the explanation is: When you mutate variables it implicitly creates an ordering dependency - later uses of the variable rely on previous mutations. However, this is an implicit dependency that isn't modeled by the language so reordering won't cause any errors.

With a very basic concrete example:

x = 7

x = x + 3

x = x / 2

Vs

x = 7

x1 = x + 3

x2 = x1 / 2

Reordering the first will have no error, but you'll get the wrong result. The second will produce an error if you try to reorder the statements.

Another way to look at it is that in the first example, the 3rd calculation doesn't have "x" as a dependency but rather "x in the state where addition has already been completed" (i.e. it's 3 different x's that all share the same name). Doing single assignment is just making this explicit.

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skeezyjefferson ◴[] No.45771110[source]
whats the difference between immutable and constant, which has been in use far longer? why are you calling it mutable?
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inanutshellus ◴[] No.45771248[source]
"Constant" implies a larger context.

As in - it's not very "constant" if you keep re-making it in your loop, right?

Whereas "immutable" throws away that extra context and means "whatever variable you have, for however long you have it, it's unchangeable."

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skeezyjefferson ◴[] No.45773019[source]
> As in - it's not very "constant" if you keep re-making it in your loop, right?

you cant change a constant though

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veilrap ◴[] No.45773195[source]
He’s implying that the variable it’s being defined within the loop. So, constant, but repeatedly redefined.
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ghurtado ◴[] No.45773473[source]
That's the opposite of what any reasonable engineer means by "constant".
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1. davrosthedalek ◴[] No.45774230[source]
No? It has a lifetime of one loop duration, and is constant during that duration. Seems perfectly fine to me.