File: primes.sh 1 #!/bin/sh 2 3 # The MIT License (MIT) 4 # 5 # Copyright (c) 2026 pacman64 6 # 7 # Permission is hereby granted, free of charge, to any person obtaining a copy 8 # of this software and associated documentation files (the "Software"), to deal 9 # in the Software without restriction, including without limitation the rights 10 # to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 11 # copies of the Software, and to permit persons to whom the Software is 12 # furnished to do so, subject to the following conditions: 13 # 14 # The above copyright notice and this permission notice shall be included in 15 # all copies or substantial portions of the Software. 16 # 17 # THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 18 # IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 19 # FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 20 # AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 21 # LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 22 # OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 23 # SOFTWARE. 24 25 26 # primes [options...] [count...] 27 # 28 # Find the first n prime numbers, emitting each one on its own line, with a 29 # default of 100 thousand primes when not given a count. 30 # 31 # The help option is `-h`, `--h`, `-help`, or `--help`. 32 33 34 case "$1" in 35 -h|--h|-help|--help) 36 awk '/^# +primes /, /^$/ { gsub(/^# ?/, ""); print }' "$0" 37 exit 0 38 ;; 39 esac 40 41 [ "$1" = '--' ] && shift 42 43 awk ' 44 BEGIN { 45 left = 100 * 1000 46 if (ARGV[1] ~ /^0+/ || ARGV[1] + 0 != 0) { 47 left = ARGV[1] + 0 48 delete ARGV[1] 49 } 50 51 if (left < 0) left = 0 52 53 printf "showing first %d prime numbers\n", left > "/dev/stderr" 54 55 if (left > 0) { 56 print 2 57 left-- 58 } 59 60 for (n = 3; left > 0; n += 2) { 61 if (positive_odd_is_prime(n)) { 62 print n 63 left-- 64 } 65 } 66 67 exit 68 } 69 70 function positive_odd_is_prime(n, max, div) { 71 max = sqrt(n) 72 for (div = 3; div <= max; div += 2) { 73 if (n % div == 0) { 74 return 0 75 } 76 } 77 return 1 78 } 79 ' "$@"