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copied!<p>Here's the rule for function sub-typing:</p> <p><strong>Argument types must be contra-variant, return types must be co-variant.</strong></p> <p>Co-variant == preserves the "A is a subtype of B" hierarchy for the type of the results parameter. </p> <p>Contra-variant == <em>reverses</em> ("goes against") the type hierarchy for the arguments parameter.</p> <p>So, in your example:</p> <pre><code>f1: int -> bool f2: bool -> bool </code></pre> <p>We can safely conclude that f2 is a subtype of f1. Why? Because (1) looking at just the argument types for both functions, we see that the type hierarchy of "bool is a subtype of int" is in fact co-variant. It preserves the type hierarchy between ints and bools. (2) looking at just the results types for both functions, we see that contra-variance is upheld.</p> <p>Put another way (the plain English way I think about this subject):</p> <p>contra-variant arguments: "my caller can pass in <em>more</em> than I require, but that's okay, because I'll use only what I need to use." co-variant return values: "I can return <em>more</em> than the caller requires, but that's okay, he/she will just use what they need, and will ignore the rest"</p> <p>Let's look at another examples, using structs where everything is an integer:</p> <pre><code>f1: {x,y,z} -> {x,y} f2: {x,y} -> {x,y,z} </code></pre> <p>So here again, we're asserting that f2 is a subtype of f1 (which it is). Looking at the argument types for both functions (and using the < symbol to denote "is a subtype of"), then if f2 < f1, is {x,y,z} < {x,y} ? The answer is yes. {x,y,z} is co-variant with {x,y}. i.e. in defining the struct {x,y,z} we "inherited" from the {x,y} struct, but added a third member, z.</p> <p>Looking at the return types for both functions, if f2 < f1, then is {x,y} > {x,y,z}? The answer again is yes. (See above logic).</p> <p>Yet a third way to think about this, is to assume f2 < f1, then try various casting scenarios, and see if everything works. Example (psuedo-code):</p> <pre><code> F1 = f1; F2 = f2; {a,b} = F1({1,2,3}); // call F1 with a {x,y,z} struct of {1,2,3}; This works. {a,b,c} = F2({1,2}); // call F2 with a {x,y} struct of {1,2}. This also works. // Now take F2, but treat it like an F1. (Which we should be able to do, // right? Because F2 is a subtype of F1). Now pass it in the argument type // F1 expects. Does our assignment still work? It does. {a,b} = ((F1) F2)({1,2,3}); </code></pre>

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