Question

I’m writing a divides-by-three function in Bash, and it won’t let me set a variable to a number.

fizzy.sh:

#!/usr/bin/env sh

div3() {
    return `$1 % 3 -eq 0`
}

d=div3 1
echo $d

Example:

$ ./fizzy.sh 
./fizzy.sh: line 7: 1: command not found

Answer 1

Bash functions normally “return” values by printing them to standard output, where the caller can capture them using

`func args ...`

or

$(func args ...)

This makes functions work like external commands.

The return statement, on the other hand, sets the value of $?. Normally that’s going to be set to 0 for success, 1 for failure, or some other value for a specified kind of failure. Think of it as a status code, not a general value. It’s likely that this will only support values from 0 to 255.

Try this instead:

#!/bin/sh

div3() {
    expr $1 % 3 = 0   # prints "0" or "1"
}

d=$(div3 1)
echo $d

Note that I’ve also changed the shebang line from #!/usr/bin/env sh to #!/bin/sh. The #!/usr/bin/env trick is often used when invoking an interpreter (such as perl) that you want to locate via $PATH. But in this case, sh will always be in /bin/sh (the system would break in various ways if it weren’t). The only reason to write #!/usr/bin/env sh would be if you wanted to use whatever sh command happens to appear first in your $PATH rather than the standard one. Even in that case you’re probably better of specifying the path to sh directly.