How to compare a program's version in a shell script?

I don't know if it is beautiful, but it is working for every version format I know.

#!/bin/bash
currentver="$(gcc -dumpversion)"
requiredver="5.0.0"
 if [ "$(printf '%s\n' "$requiredver" "$currentver" | sort -V | head -n1)" = "$requiredver" ]; then 
        echo "Greater than or equal to ${requiredver}"
 else
        echo "Less than ${requiredver}"
 fi

(Note: better version by the user 'wildcard': https://unix.stackexchange.com/users/135943/wildcard , removed additional condition)

Shorter version:

version_greater_equal()
{
    printf '%s\n%s\n' "$2" "$1" | sort -V -C
}

version_greater_equal "${gcc_version}" 8.2 || die "need 8.2 or above"

Here I give a solution for comparing Unix Kernel versions. And it should work for others such as gcc. I only care for the first 2 version number but you can add another layer of logic. It is one liner and I wrote it in multiple line for understanding.

check_linux_version() {
    version_good=$(uname -r | awk 'BEGIN{ FS="."}; 
    { if ($1 < 4) { print "N"; } 
      else if ($1 == 4) { 
          if ($2 < 4) { print "N"; } 
          else { print "Y"; } 
      } 
      else { print "Y"; }
    }')

    #if [ "$current" \< "$expected" ]; then
    if [ "$version_good" = "N" ]; then
        current=$(uname -r)
        echo current linux version too low
        echo current Linux: $current
        echo required 4.4 minimum
        return 1
    fi
}

You can modify this and use it for gcc version checking.