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AgileSoul
Asked: 2022-07-23 15:52:37 +0800 CST 2022-07-23 15:52:37 +0800 CST

What does %02X $((0x${MAC_ADDRESS:13:1} + 1)) mean in bash using printf?

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I'm trying to understand this code 100% but I'm not completely clear, especially because of the printf %02X:

MAC_ADDRESS="AF:BA:44:2A:8F:3C"
# Se que sacamos un dijito a la derecha de la posicion '13' con ${MAC_ADDRESS:13:1}
printf %02X $((0x${MAC_ADDRESS:13:1} + 1))
#output:
10

I know that addition of hexadecimal numbers is being allowed, but I don't know what allows it or the purpose of it %02Xas the first argument to the command printf.

# Algo raro es que al querer sumar "1 + D" me da un resultado erroneo, pues el resultado real deberia ser "0E"
printf %02X $((0x1+D))
#output:
01
bash

1 Answers

  1. Best Answer

    Let's go by parts. Let's divide everything that happens for printfinto three parts:

    • %02X formats output as two hexadecimal digits.
    • ((expression)) everything in there evaluates to an arithmetic expression.
    • ${string:pos:characters} extracts n characters from the substring from pos and appends it to 0x

    Therefore you will have printf %02X $((0xF+1)), which in hexa is 10

    Regarding your doubt when doing $((0x1+D)), you are adding the letter D to 0x1 (hexadecimal).

    Maybe you wanted to do the following$((0xD+1))

    I hope it helps you, greetings!

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