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The equation "(f >=> g) x = (f(x) → mb) >>= g(y = b)" is very confusing, it's unclear where mb and y are bound; I have never seen anything similar. The explanation in this thread is also misguided: x is not a type and there is no substitution happening that forms functors.
It makes sense to avoid Haskell-specific notation, but here the only notation used is function application written as "f x", and binary operator "f >=> g". I restored the equation, using f(x) as function application. That should be understandable without additional prerequisites. [[Special:Contributions/2001:861:3F42:1B60:AEFC:B99:1842:A29E|2001:861:3F42:1B60:AEFC:B99:1842:A29E]] ([[User talk:2001:861:3F42:1B60:AEFC:B99:1842:A29E|talk]]) <!--Template:Undated--><small class="autosigned">— Preceding [[Wikipedia:Signatures|undated]] comment added 07:57, 24 November 2022 (UTC)</small> <!--Autosigned by SineBot-->
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