trying to create printf script

Shell Programming and Scripting
1

I am trying to make a script that randomly generates 4 sets of numbers with decimal points. Then it outputs that to a variable and a file. This is what I have:

#!/bin/bash

printf -v RANDOM_2 "%s\n" "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM"
printf "%s\n" "$RANDOM_2" > ./bp1/bin/random

For some reason when I do these commands separate in terminal it works fine but together in a script it wont work right. Anyone know how to make this work?

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2

You're using printf like echo, which it's not. 'printf "%s" a b c d' will only print a because you're only telling it to print one string with '%s'. You might as well just use echo and print all of them.

You don't even need a variable here. Just echo it right into the file.

echo "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM" "$RANDOM"."$RANDOM" > ./bp1/bin/random

This may not be the best way to generate random numbers. If RANDOM is between 0 and 65536, you'll never have a decimal point higher than .65. Let's try printf again, properly, with:

#!/bin/bash
( for ((N=0; N<4; N++))
do
        printf " %d.%03d" "$((RANDOM%1000))" "$((RANDOM%1000))"
done ) > ./bp1/bin/random

This should generate random numbers between 0 and 1000, with three decimal places.

3

I know this isn't the proper way to do it but I am currently in a Linux/Unix programming class and they want this to be done using printf and $RANDOM (which I've never done it this way before). I need a new variable and a file both with the same outputted data.

4

Actually, that will print out 'abcd' (without a trailing newline). printf reuses the format string until all arguments are exhausted.

Example:

$ printf 'Line %d: %s\n' 1 one 2 two 3 three
Line 1: one
Line 2: two
Line 3: three

Regards,
Alister