Host introspection
You are encouraged to solve this task according to the task description, using any language you may know.
Print the word size and endianness of the host machine.
C
<c>#include <stdio.h>
int main() {
int word = 1;
printf("word size = %d\n", 8 * sizeof(void *)); // a pointer takes up a word
if (*(char *)&word) // if the least significant bit is located in the lowest-address byte
printf("little endian\n");
else
printf("big endian\n");
return 0;
}</c>
Forth
: endian cr 1 cells . ." address units per cell" s" ADDRESS-UNIT-BITS" environment? if cr . ." bits per address unit" then cr 1 here ! here c@ if ." little" else ." big" then ." endian" ;
This relies on c@ being a byte fetch (4 chars = 1 cells). Although it is on most architectures, ANS Forth only guarantees that 1 chars <= 1 cells. Some Forths like OpenFirmware have explicitly sized fetches, like b@.
J
":&> (|: 32 64 ;"0 big`little) {"_1~ 2 2 #: 16b_e0 + a. i. 0 { 3!:1 ''
32
little
OCaml
Print word size: <ocaml>Printf.printf "%d\n" Sys.word_size</ocaml> Dunno about endianness