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> The only thing that makes Lisp unique in this context is that it also makes arithmetic operations prefix I think that most problematic for prefix notation ar

> The only thing that makes Lisp unique in this context is that it also makes arithmetic operations prefix I think that most problematic for prefix notation are not arithmetic operations, but field accessors. In C you have a->b->c->d, in Lisp it would be (d (c (b a))), which makes you to jump to the center and read it from the inside-out.

lispm 2y ago
In Lisp one could have (-> a b c d), just like (+ (+ 2 3) 4) is (+ 2 3 4).
Zambyte 2y ago
The direct translation of the C expression to Lisp would actually be (-> a (-> b (-> c d))), which is fine to read from left to right as well. As lispm said, this can be reduced to (-> a b c d) due to the flexibility of prefix operations. The direct translation of the Lisp expression you shared to C would be: d(c(b(a))), which just like the Lisp expression, evaluates from the middle out. Both are fine to read from left to right though.
- zajio1am 2y ago
  > The direct translation of the C expression to Lisp would actually be (-> a (-> b (-> c d))) Not really. -> looks like binary infix operator, but it is really an unary postfix operator (parametrized by field). Because field itself is not a first-class entity in C.
  - Zambyte 2y ago
    Either -> is a binary operator or is is not an operator in the expression at all, and the unary postfix operators are ->b, ->c, and ->d. In the latter case, that gives us (->d (->c (->b a))) in Lisp, which I can see how ypu could get, but I don't agree. #include <stdio.h> #include <stdlib.h> struct foo { int a; }; int main() { struct foo *f = malloc(sizeof(struct foo)); f->a = 5; printf("%d\n", f->b); free(f); return 0; } test.c:11:21: error: no member named 'b' in 'struct foo' 11 | printf("%d\n", f->b); | ~ ^ 1 error generated. What do you mean when you say fields are not first class in C?

lispm 2y ago

In Lisp one could have (-> a b c d), just like (+ (+ 2 3) 4) is (+ 2 3 4).

Zambyte 2y ago

The direct translation of the C expression to Lisp would actually be (-> a (-> b (-> c d))), which is fine to read from left to right as well. As lispm said, this can be reduced to (-> a b c d) due to the flexibility of prefix operations. The direct translation of the Lisp expression you shared to C would be: d(c(b(a))), which just like the Lisp expression, evaluates from the middle out. Both are fine to read from left to right though.

zajio1am 2y ago
> The direct translation of the C expression to Lisp would actually be (-> a (-> b (-> c d))) Not really. -> looks like binary infix operator, but it is really an unary postfix operator (parametrized by field). Because field itself is not a first-class entity in C.
- Zambyte 2y ago
  Either -> is a binary operator or is is not an operator in the expression at all, and the unary postfix operators are ->b, ->c, and ->d. In the latter case, that gives us (->d (->c (->b a))) in Lisp, which I can see how ypu could get, but I don't agree. #include <stdio.h> #include <stdlib.h> struct foo { int a; }; int main() { struct foo *f = malloc(sizeof(struct foo)); f->a = 5; printf("%d\n", f->b); free(f); return 0; } test.c:11:21: error: no member named 'b' in 'struct foo' 11 | printf("%d\n", f->b); | ~ ^ 1 error generated. What do you mean when you say fields are not first class in C?