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Home/ Questions/Q 7709105
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T00:46:52+00:00 2026-06-01T00:46:52+00:00

When using a computational expression, the first definition works but the second does not

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When using a computational expression, the first definition works but the second does not for Zero.

What is the difference between this:

member o.Zero() = 3

and this:

member o.Zero = fun() -> 3

The first evaluates to unit -> int and the second to (unit -> int). What is the difference?

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  1. Editorial Team
    Editorial Team
    2026-06-01T00:46:54+00:00Added an answer on June 1, 2026 at 12:46 am

    If they are let-bounds, there is no difference. However, in a class definition the first o.Zero is a method while the second o.Zerois a property.

    Computation expression expects a method named Zero; that’s why it didn’t work when you provided a property with the same name.

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