#!/bin/sh

# The MIT License (MIT)
#
# Copyright © 2024 pacman64
#
# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the “Software”), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
#
# The above copyright notice and this permission notice shall be included in
# all copies or substantial portions of the Software.
#
# THE SOFTWARE IS PROVIDED “AS IS”, WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
# SOFTWARE.


# primes [options...] [count...]
#
# Find the first n prime numbers, emitting each one on its own line, with a
# default of 100 thousand when not given a count.
#
# The help option is `-h`, `--h`, `-help`, or `--help`.


# handle leading options
case "$1" in
    -h|--h|-help|--help)
        awk '/^# +primes/, /^$/ { gsub(/^# ?/, ""); print }' "$0"
        exit 0
    ;;
esac

awk '
BEGIN {
    left = 100 * 1000
    if (ARGV[1] + 0 != 0) {
        left = ARGV[1] + 0
        delete ARGV[1]
    }

    if (left < 0) {
        left = 0
    }

    printf "showing first %d prime numbers\n", left > "/dev/stderr"

    if (left > 0) {
        print 2
        left--
    }

    for (n = 3; left > 0; n += 2) {
        if (is_prime(n)) {
            print n
            left--
        }
    }

    exit
}

function is_prime(n, max, div) {
    if (n < 2) {
        return 0
    }

    if (n % 2 == 0) {
        return n == 2
    }

    max = sqrt(n)
    for (div = 3; div <= max; div += 2) {
        if (n % div == 0) {
            return 0
        }
    }
    return 1
}
' "$@"