Let’s say I want to write the factorial function in Julia in the following recursive way: FACTORIAL(n) = n==1 ? 1 : n*BigInt(FACTORIAL(n-1)) Doing so, it shows the following error when trying to compute FACTORIAL(1e6) julia> FACTORIAL(1e6) ERROR: StackOverflowError: Stacktrace: [1] FACTORIAL(n::…

At first I thought this was a type issue, or comparison issue in the base case. But it’s not. The reason is you’ve recursively called factorial 65428 times and it’s overflowed the stack. Unlike scheme for example there is no tail recursion elimination in Julia, for good reasons to do with multiple …

A “ stack overflow ” is not the same thing as “numerical overflow.” A stack overflow is usually caused by a recursive function that recurses “too many” times. Every time it calls itself, it has to add its data to the top of the stack (a part of memory designated for basic program data). The stack is p…

[image] mikmoore: Some languages utilize “tail call optimization” to eliminate the excess memory use of some simple recursive functions Even if Julia had TCO, it wouldn’t work for the FACTORIAL(n) function here because the recursive call is not in tail position. (This seems to be a widespre…

[image] dlakelan: Unlike scheme for example there is no tail recursion elimination in Julia, for good reasons to do with multiple dispatch. The stated reason has nothing to do with multiple dispatch, and is one I only halfway agree with. The comment itself is on point, as is the solution. Wh…

As @stevengj pointed out though, in this case tail-call elimination wouldn’t even help because the recursive call isn’t in tail position.

Why do Julia BigInt iterative functions overflow?

General Usage

stevengj February 23, 2024, 11:47pm 8

Please, not another TCO thread. See e.g. Does Julia have tail call optimization? and references therein. There is nothing new to say on this topic.

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Why do Julia BigInt iterative functions overflow?

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