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Go has GC and Rust doesn't. If you were to design an explicitly C-like language with memory safety and without GC, it would look a lot more like Rust than Go. The idea that Rust was designed by a bunch of C++ fans is absurd. Most of us were ML fans more than anything. t.co/SWJhLJptY3

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Patrick Walton

@pcwalton

Jul 25, 2019

The issue is that generics are basically a requirement for memory safety without GC as Rust does it. Otherwise you can't create safe abstractions, which would mean that every time you want a linked list you have to write unsafe code. That would undermine memory safety.

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I don't follow. Do you claim this is the case if you did it the C way, i.e. without an abstraction or with macros?

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How do you write a function receiving a slice like &[Type] and returning a reference to the initial value as &Type if you don't have generics? It needs to be generic based on the lifetime of the reference even if Type is a specific type rather than a generic type parameter.

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I don't understand Rust's model for this so I don't know.

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You need a way for the language to understand that the lifetime of the returned reference is based on the lifetime of the received reference. For non-trivial cases, the language needs a way to express what lifetimes are based on. Lifetimes are part of types so you need generics.

9:06 PM · Jul 25, 2019Twitter Web App

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and

A lifetime is the scope where something is valid, either the global scope or a lexical block in a function. For verifying correctness, the type checker is more clever than only using lexical scopes (doc.rust-lang.org/edition-guide/) but they're fundamentally based on lexical scopes.

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In order for the language to be usable, you need a way to write code that's generic across different lifetimes along with a way to express which lifetime something is based on for non-trivial cases where the language can implicitly infer what is meant. It's still generic though.