If it implements a subset of a scheme standard then it is not an implementation of a scheme standard. Is that not obvious?

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kazinator 7 months ago | parent | next [-]

Something that implements a subset of Scheme is not a complete implementation of the Scheme specification, but to the extent that it is complete, mit is a Scheme dialect.

There are Scheme programs which the implementation handles, and programs developed with that incomplete implementation can be run by more complete implementations.

A Scheme programmer informed on the limits of the subset (what is not in it) implementation can jump in and start programming using what is in the subset.

A subset language can be useful to practitioners of the full language.

ThinLisp is an incomplete implementation of Common Lisp, but one which compiles to C that doesn't need garbage collection. You can develop ThinLisp programs as Common Lisp programs (with certain care). Then ThinLisp translates them to C. I think that (by default?) these translated programs don't need garbage collection.

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Pet_Ant 7 months ago | parent | prev [-]

I can't tell if you are joking or not.

If a platypus lays eggs, but it meets other criteria of being a mammal, it's still called a mammal. Owl may not be a full conforming Scheme implementation, but it seems to be member of the Scheme sub-family.

I'd say if you are a Lisp-1 (ie one namespace for variables and functions) and generally use the function names used by the Scheme standard then you are a Scheme (or Scheme-like-Lisp if you are very taxonomically retentive). However, if you are a Lisp-2 and use the function names of Common Lisp you're a Lisp.

Maybe if one were doing a deep dive on the history of Lisp and are into S-expressions vs M-expressions then it might be too loose, but I believe what is above is what most people would agree with in this context.

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binary132 7 months ago | parent [-]

Lisp is a very general category (I would consider Clojure a Lisp, with some extras) while Scheme is not, it is defined by a specification. If some implementation does not implement the entire specification of a given protocol, it doesn't implement that protocol, right? Would a C++ compiler that doesn't accept "new" and "delete" be C++? No, it's some subset of C++, not C++. And some implementation of HTTP that doesn't have any verbs wouldn't be HTTP. An implementation of a Scheme specification that doesn't implement the full specification is not a Scheme. No?

A mammal is a creature that has X, Y, and Z characteristics. A creature having only X and Y characteristics is therefore not a mammal.

	▲	lispm 7 months ago \| parent \| next [-]
		Scheme is a general category of probably 100+ implementations of various versions of the language based on various specs and standards. Small educational languages based on R4RS, R5RS. Then there is the more controversial R6RS and later a R7RS small and attempts to define a R7RS large. Then there are 100+ of language extensions in various states described in SRFIs. The there is an official language standard IEEE Scheme, which is not widely used (AFAIK). Then there are various subsets of the language Scheme used in education. For example in SICP there is a minimal Scheme subset used in the book. Scheme also spawned its own dialects (or maybe derived language), like Racket, which then has a new dialect Rhombus... Lisp is either a family of language families (Lisps, Scheme, Logo, ML, MDL, JavaScript, Dylan, Racket, R, ...) or a more narrowly defined family of languages (often recognizable because they identify as Lisp in their name) and based on the core from Lisp 1: Lisp 1, Lisp 1.5, BBN Lisp, Interlisp, Maclisp, Franz Lisp, ZetaLisp, Le_Lisp, EuLisp, muLisp, Common Lisp (a standardized language), Emacs Lisp, Standard Lisp, ISLISP (a standardized language), ... You'll find often in the Scheme literature Lisp described as a different, but related/derived, language (more symbolic, less functional, more procedural, procedural macros, less clean, less systematic, more dynamic, ...). Scheme is seen more functional, more lexical binding, less procedural, more minimal, one namespace, cleaner, ...).
	▲	kazinator 7 months ago \| parent \| prev [-]
		> Would a C++ compiler that doesn't accept "new" and "delete" be C++ Believe it or not, yes it would. It would not be a conforming implementation of any ISO C++ dialect, but it would be recongizable as a C++ dialect and mutually intelligible. Someone who knows C++ would understand programs written in that dialect, and could write new ones in it. Something which is not a conforming ISO C++ implementation probably has good company; most production compilers are likely not fully conforming. Just not in gaping ways like new and delete being absent.