Bash script that accepts a number from the command line and performs functions until the number is 1

Question

I am trying to make a bash script that takes in a number from command line (between 1 and 50). If the number is even then it divides the number by 2, and if the number is odd then it multiplies it by 3 and adds 1. The number then goes back through the functions until the number 1 is reached. Here is what I have so far

#!/bin/bash

//this looks to see if the command line argument, $1, is between 1 and 50
if ["$1" -gt 1 ] && [ "$1" -lt 50]
then
//if $1 is between 1 and 50, it goes into a for loop when it looks to see if it is even or odd. It keeps going until the number 1 is reached, and prints out each number until then
for i in $1; do
    //if $1 is even
    if [i%2 -eq 0]
    then
        i/2 == i  
        echo "i,"
    //if $1 is even
    else 
        3*i+1 == i
        echo "i,"
    fi
done
fi 

When I run the code, I get an error.

$ ./q1.sh 5

./q1.sh: line 3: [5: command not found

I have looked online to see what the error means but cant find an answer. Can someone please help me.

Answer 1

Just looking at your requirements, I would solve this as follows:

#!/bin/bash

# Assign positional parameter
num="$1"

if (( num > 1 && num < 50 )); then
    while (( num != 1 )); do
        if (( num % 2 == 0 )); then # Number is even
            (( num /= 2))           # Compound assignment: same as num = num / 2
        else                        # Number is odd
            (( num = num * 3 + 1 ))
        fi
        printf "Number is %d\n" "$num"
    done
fi

This makes heavy use of arithmetic expressions (( ... )) for conditions and calculations. In fact, I don't use the test construct [ ... ] ever. If I use this, I can even assign within the expression with, for example, (( num = num * 2 )) instead of num=$(( num * 2 )).

Notice how the while loop checks if we have reached the value 1 yet. The way you tried it, with for i in $1, is just a (complicated) way to assign $1 to i, but the loop is executed just once.

Things from your code that won't work

• Comments start with #, not with //.
• The test construct [ ... ] requires spaces: [ "$1" -gt 1 ] etc.
• The for loop just assigns $1 to i, nothing else.
• [ i%2 -eq 0 ] compares the literal i%2, and not the result of the variable i modulo 2; you'd need [ $(( i%2 )) -eq 0 ] for that, or (( i%2 == 0 )).
• The assignment i/2 == i will try to execute the command i/2 with the arguments == and i – with little success. You want i=$(( i/2 )) or (( i /= 2 )).
• echo "i," just prints i,. You want echo "$i,".
• For the assignment 3*i+1 == i, see above.

Answer 2

In addition to missing a spaces in [ i%2 -eq 0 ] and [ "$1" -lt 50 ], your arithmetic use is incorrect along with several variable assignment issues. You need to assign the result of the arithmetic to the value at hand. The form is var=value, not value == var.

You have 3 choices for how to do arithmetic in bash. The old newval=$(expr i / 2) or let newval="$i"/2 or newval=$((i / 2)):

Cleaning up your script, you could do something like:

if [ "$1" -gt 1 ] && [ "$1" -lt 50 ]
then
    for i in $1; do
        //if $i is even
        if (( i % 2 == 0 ))
        then
            newvar=$((i / 2))
            echo "$newvar,"
        //if $i is odd
        else 
            newvar=$((3 * i + 1))
            echo "$newvar,"
        fi
    done
fi

Note, while you can multiply and reassign the loop increment i, it is far better to assign the result to a new variable, e.g. some newvar.

Note how:

newvar=$((i / 2))

and

newvar=$((i * 3 + 1))

It also looks like you intended:

if [ "${#1}" -gt 1 ] && [ "${#1}" -lt 50 ]

to check that the number of characters given is between 1 and 50, otherwise, your for i in $1; do does not make a lot of sense. If you are just testing the value and not the number of characters, get rid of the for altogether:

if [ "$1" -gt 1 -a "$1" -lt 50 ]
then
    val="$1"              ## assign positinal parameter to variable
    ## if $val is even    ## to avoid subtleties in syntax
    if (( val % 2 == 0 ))
    then
        newvar=$((val / 2))
        echo "$newvar,"
    ## if $val is odd
    else 
        newvar=$((3 * val + 1))
        echo "$newvar,"
    fi
fi

Alternative

If I have figured out what you are trying to do, then you might consider doing it this way:

#!/bin/bash

declare -i val
if [ "$1" -gt 1 -a "$1" -lt 50 ]; then
    val="$1"                    # assign val=$1
    while ((val > 0)); do       # loop for val > 0
        if ((val > 1)); then    # set output format
            fmt="%d,"
        else
            fmt="%d"
        fi
        if ((val % 2 == 0))     # if $val is even
        then
            n=$((val / 2))
        else                    # if $val is even
            n=$((val * 3 + 1))
        fi
        printf "$fmt" "$n"
        ((val--))               # decrement val
    done
else
    printf "input value not between 1 and 50\n"
fi

printf "\n"

Example Use/Output

$ bash ql.sh 10
5,28,4,22,3,16,2,10,1,4

Or, rewritten in a more compact form using compound commands:

#!/bin/bash

declare -i val
if [ "$1" -gt 1 -a "$1" -lt 50 ]; then          # validate input 1 - 50
    for ((val = $1; val > 0; val--)); do        # loop for val > 0
        ((val > 1)) && fmt="%d," || fmt="%d"    # print format
        ((val % 2 == 0)) && n=$((val / 2)) || n=$((val * 3 + 1)) # even/odd
        printf "$fmt" "$n"
    done
else
    printf "input value not between 1 and 50\n"
fi

printf "\n"