4 ms·
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, th
by 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 agoEither -> 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?