4 ms·

how do you not use bounds checking? (unsafe like get_unchecked doesn't count because then you can say that about almost every rust feature).

deleted 7y ago
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steveklabnik 7y ago
You're mis-understanding what "you don't pay for what you don't use" means. It means that a program that does not contain [] in it does not include code for []. There's no "mandatory runtime" code for []. It doesn't mean you can use [] in some special way and somehow get less code. And even if we were talking about your understanding, I don't see why get_unchecked would be disqualified. unsafe is a part of the language for a reason.
- jnordwick 7y ago
  I know, the meaning of "zero cost abstraction" is really "zero cost to not use the abstraction." But that's like saying if you don't use rust you don't pay for it. Just because there is the unsafe escape hatch in the language, you don't get to label the language as zero cost abstraction. Because practically speaking there is a lot more runtime cost to rust than people will tell you. Many langauges (almost anything without a runtime) meet that definition of zero cost abstraction then.
  - jzoch 7y ago
    What languages dont have a runtime? Even C has one (albeit a very small one). Nobody labels Rust the language as a zero cost abstraction (thatd be silly - there is a cost to learning it!). Rather, they try to provide zero cost abstractions. A great example is that there are no green threads in rust because they consciously removed them as they penalized rust performance regardless of whether said people used green threads or not.
  - lilyball 7y ago
    You still seem to be missing the point. If you use a Rust construct that does bounds-checked accesses, you're explicitly using bounds checks. You're not paying for anything beyond the bounds checks that you're using. If you use a Rust construct that elides the bounds checks (e.g. get_unchecked), there are no bounds checks, and you don't pay any cost for the fact that Rust supports a bounds-checked subscript operator. If you want to compare Rust's subscript operator, do so with an explicitly bounds-checked version of the C code. That's the appropriate comparison for "zero-cost abstraction". Because the whole point of "abstraction" is it's not a change to the observed runtime behavior, it's just a language abstraction.
    - jnordwick 7y ago
      What part of my comment made you think I didn't understand that? Was it when I wrote: "it really [is] zero cost to not use the abstraction"? Or when I wrote: "if you don't use rust you don't pay for it"? I understand the concept fine. It is rust's marketing gimmick and not a useful tool to describe the language. Rust is no more "zero cost" than C++, Fortran, Ada, Objective C, and even D can even make the claim to some extent (you can opt out of the GC and use malloc/free directly). I'm there are plenty more I'm missing. If you use the Java GC that doesn't collect (used for real-time programs), even that probably fits the description. Under your description an opt in generational GC is zero cost. You can't have the vector be checked, (along with other small performance issues), but use the excuse that you can write code in a way that doesn't use checked access or have that feature's cost. I can do that in a lot of languages. Also to some extent you should take the ecosystem into consideration where a large chunk of it goes to lengths to not use unsafe such as using a vector for homogeneous objects to get around pointer semantics. That incurs a huge cost, and that should be counted against the language because it makes the other ways too difficult or the main libraries used rely on those techniques. (and if you don't count rust's ersatz standard library, the you can't count Java's standard library since you can always write Java code that doesn't use a GC or extra features - I've done it a few times). I saw so much promise in rust, but is seems to have gone to complete crap.
      - Tomte 7y ago
        > Rust is no more "zero cost" than C++ C++ popularized the term. And if C++ is commonly described as having designed its features to be zero-cost (or you only pay for what you choose), there's nothing wrong with Rust describing the same concept with the same term.

deleted 7y ago

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steveklabnik 7y ago

You're mis-understanding what "you don't pay for what you don't use" means. It means that a program that does not contain [] in it does not include code for []. There's no "mandatory runtime" code for []. It doesn't mean you can use [] in some special way and somehow get less code. And even if we were talking about your understanding, I don't see why get_unchecked would be disqualified. unsafe is a part of the language for a reason.

jnordwick 7y ago
I know, the meaning of "zero cost abstraction" is really "zero cost to not use the abstraction." But that's like saying if you don't use rust you don't pay for it. Just because there is the unsafe escape hatch in the language, you don't get to label the language as zero cost abstraction. Because practically speaking there is a lot more runtime cost to rust than people will tell you. Many langauges (almost anything without a runtime) meet that definition of zero cost abstraction then.
- jzoch 7y ago
  What languages dont have a runtime? Even C has one (albeit a very small one). Nobody labels Rust the language as a zero cost abstraction (thatd be silly - there is a cost to learning it!). Rather, they try to provide zero cost abstractions. A great example is that there are no green threads in rust because they consciously removed them as they penalized rust performance regardless of whether said people used green threads or not.
- lilyball 7y ago
  You still seem to be missing the point. If you use a Rust construct that does bounds-checked accesses, you're explicitly using bounds checks. You're not paying for anything beyond the bounds checks that you're using. If you use a Rust construct that elides the bounds checks (e.g. get_unchecked), there are no bounds checks, and you don't pay any cost for the fact that Rust supports a bounds-checked subscript operator. If you want to compare Rust's subscript operator, do so with an explicitly bounds-checked version of the C code. That's the appropriate comparison for "zero-cost abstraction". Because the whole point of "abstraction" is it's not a change to the observed runtime behavior, it's just a language abstraction.
  - jnordwick 7y ago
    What part of my comment made you think I didn't understand that? Was it when I wrote: "it really [is] zero cost to not use the abstraction"? Or when I wrote: "if you don't use rust you don't pay for it"? I understand the concept fine. It is rust's marketing gimmick and not a useful tool to describe the language. Rust is no more "zero cost" than C++, Fortran, Ada, Objective C, and even D can even make the claim to some extent (you can opt out of the GC and use malloc/free directly). I'm there are plenty more I'm missing. If you use the Java GC that doesn't collect (used for real-time programs), even that probably fits the description. Under your description an opt in generational GC is zero cost. You can't have the vector be checked, (along with other small performance issues), but use the excuse that you can write code in a way that doesn't use checked access or have that feature's cost. I can do that in a lot of languages. Also to some extent you should take the ecosystem into consideration where a large chunk of it goes to lengths to not use unsafe such as using a vector for homogeneous objects to get around pointer semantics. That incurs a huge cost, and that should be counted against the language because it makes the other ways too difficult or the main libraries used rely on those techniques. (and if you don't count rust's ersatz standard library, the you can't count Java's standard library since you can always write Java code that doesn't use a GC or extra features - I've done it a few times). I saw so much promise in rust, but is seems to have gone to complete crap.
    - Tomte 7y ago
      > Rust is no more "zero cost" than C++ C++ popularized the term. And if C++ is commonly described as having designed its features to be zero-cost (or you only pay for what you choose), there's nothing wrong with Rust describing the same concept with the same term.