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Home/ Questions/Q 590335
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T15:31:07+00:00 2026-05-13T15:31:07+00:00

I was just having a quick read through this article (specifically the bit about

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I was just having a quick read through this article (specifically the bit about why he chose to use structs / fields instead of classes / properties) and saw this line:

The result of a property is not a true l-value so we cannot do something like Vertex.Normal.dx = 0. The chaining of properties gives very unexpected results.

What sort of unexpected results is he talking about?

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  1. Editorial Team
    Editorial Team
    2026-05-13T15:31:07+00:00Added an answer on May 13, 2026 at 3:31 pm

    I would add to dbemerlin’s answer that the key here is Rico’s note that properties are not “lvalues”, or, as we call them in C#, “variables”.

    In order to mutate a mutable struct (and ideally, you should not; mutable structs often cause more problems than they solve) you need to mutate a variable. That’s what a variable is — a storage location whose contents change. If you have a field of type vector and you say

    Foo.vector.x = 123;
    

    then we have a variable of value type — the field Foo.vector — and we can therefore mutate its property x. But if you have a property of value type:

    Foo.Vector.x = 123;
    

    the property is not a variable. This is equivalent to

    Vector v = Foo.Vector;
    v.x = 123;
    

    which mutates the temporary variable v, not whatever storage location is backing the property.

    The whole problem goes away if you abandon mutable value types. To change x, make a new vector with the new values and replace the whole thing:

    Foo.Vector = new Vector(x, Foo.Vector.y);
    
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