bash function return in if statement not working

Question

I'm trying to refactor this code:

if [ $(($1 % 4)) -eq 0 ] && [ $(($1 % 100)) -ne 0 ] || [ $(($1 % 400)) -eq 0 ] ; then
    echo $T
else
    echo $F
fi

into something like this:

if divisibleBy4 && notDivisibleBy100 || divisibleBy400; then
    echo $T
else
    echo $F
fi

note that

T="true"
F="false"

divisibleBy4 function looks like:

divisibleBy4() {
    return  [ $(($1 % 4)) -eq 0 ]
}

But I've tried several iterations including what I thought would definitely work.

divisibleBy4() {
    if [ $(($1 % 4)) -eq 0  ]; then
    return 1
    else return 0
    fi
}

Any idea how to properly fix the syntax so I can refactor these into functions and use them in my if statement?

When testing I'm seeing the error

syntax error: operand expected (error token is "% 4")

Another thing I tried is, but still doesn't seem to work:

INPUT=$1

divisibleBy4() {
    if [ $(($INPUT % 4)) -eq 0  ]; then
         return 1
    else return 0
    fi
}

notDivisibleBy100() {
    if [ $(($INPUT % 100)) -ne 0]; then
         return 1
    else return 0
    fi
}

divisibleBy400() {
    if [ $(($INPUT % 400)) -eq 0  ]; then
         return 1
    else return 0
    fi
}


if divisibleBy4 && notDivisibleBy100 || divisibleBy400; then
    echo $T
else
    echo $F
fi

or

INPUT=$1

divisibleBy4() {
    return $((!($INPUT %4)))
}
notDivisibleBy100() {
    return $(($INPUT %100))
}
divisibleBy400() {
    return $((!($INPUT %400)))
}

(( divisibleBy4 && notDivisibleBy100 || divisibleBy400 )) && echo "true" || echo "false"

return is not exactly like some other language's returns. E.g.: ``` divisibleBy4() { if [ $(($1 % 4)) -eq 0 ]; then return 5; else return 17; fi } $(divisibleBy4 4) echo $? # prints 5 $(divisibleBy4 19) echo $? # prints 17 ``` — userSeventeen
– userSeventeen, Commented Jan 26, 2019 at 19:36
Your latter version works (though your logic is inverted: 0 should be success), it's just that in a function $1 refers to the function's arguments and not the script's. You'd have to call it as divisibleBy4 "$1". ShellCheck autodetects this issue. — that other guy
– that other guy, Commented Jan 26, 2019 at 19:39
@thatotherguy oh that makes sense! But if I add a var INPUT=$1 divisibleBy4() { if [ $(($INPUT % 4)) -eq 0 ]; then return 1 else return 0 fi } doesn't seem to work — Turnipdabeets
– Turnipdabeets, Commented Jan 26, 2019 at 19:50
You seem to imagine that [ is part of the shell's syntax, and/or that the argument to return can be an executable command to evaluate. Both of these are untrue. — tripleee
– tripleee, Commented Jan 26, 2019 at 20:36
@tripleee is there no way to execute a command and then use the return of that command to evaluate in an if statement? — Turnipdabeets
– Turnipdabeets, Commented Jan 26, 2019 at 20:40

Wiimm · Accepted Answer · 2019-01-26 19:50:26Z

3

You want to detect a leap year! A complete other solution using math mode directly:

a="$1"; (( !(a%4) && a%100 || !(a%400) )) && echo true || echo false

or as if-then-else

a="$1";
if (( !(a%4) && a%100 || !(a%400) )); then
    echo true
else
    echo false

answered Jan 26, 2019 at 19:50

Wiimm

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4 Comments

Turnipdabeets Over a year ago

haha you're right! That does look look cleaner - it looks like a%100 would be the opposite of what I want. I get that !(a%4) makes a 0 into a 1 but I guess if it has a remainder like a%100 then it processes as false or a 0?

chepner Over a year ago

!x is equal to 1 if x is zero, and 0 if x is any other non-zero value.

Wiimm Over a year ago

Try for ((a=1890;a<2010;a++)); do (( !(a%4) && a%100 || !(a%400) )) && echo $a; done and you see, that the expression is ok.

Turnipdabeets Over a year ago

I see, I get it thanks! Even though this might be a cleaner solution, do you have any insight on the syntax for breaking it up into separate functions so I can call if divisibleBy4 && notDivisibleBy100 || divisibleBy400;

Gordon Davisson · Accepted Answer · 2019-01-26 21:18:46Z

The simplest, directest answer is to just create functions that consist only of the tests themselves:

INPUT=$1

divisibleBy4() {
    [ $(($INPUT % 4)) -eq 0  ]
}

notDivisibleBy100() {
    [ $(($INPUT % 100)) -ne 0 ]
}

divisibleBy400() {
    [ $(($INPUT % 400)) -eq 0  ]
}

The reason this works is that a function without a return will implicitly return the status of the last command in the function; in these cases, that's the test command (note: [ is a command, even though it doesn't look like one), so the functions just return the result of the test directly.

I'd make at least one change to these, though: they all test the value of the shell variable INPUT; it's much better practice to actually pass the data that functions operate on as parameters. Thus, it'd be better to do something like this:

divisibleBy4() {
    [ $(($1 % 4)) -eq 0  ]
}

if divisibleBy4 "$1" ...

Rather than this:

divisibleBy4() {
    [ $(($INPUT % 4)) -eq 0  ]
}

INPUT=$1
if divisibleBy4 ...

Note that you can also bundle up the whole leap year check the same way:

isLeapYear() {
    [ $(($1 % 4)) -eq 0 ] && [ $(($1 % 100)) -ne 0 ] || [ $(($1 % 400)) -eq 0 ]
}

if isLeapYear "$1"; then

Or use the simpler form @Wiimm suggested:

isLeapYear() {
    (( !($1%4) && $1%100 || !($1%400) ))
}

Also, for the shell variables you do use, lower- or mixed-case is preferred, to avoid accidental conflicts with the many all-caps variable names that have special meanings or functions.

tripleee · Accepted Answer · 2019-01-27 07:22:08Z

0

Every command sets a result code. If you want to force each calculation to happen in a separate function, you can say

divisibleBy4() {
    $((!("$1" % 4)))
}
notDivisibleBy100() {
     $(("$1" %100))
}
divisibleBy400() {
     $((!("$1" %400)))
}

(divisibleBy4 "$1" &&
 notDivisibleBy100 "$1" ||
 divisibleBy400 "$1") &&
echo "true" || echo "false"

Breaking up your logic to functions on the subatomic level is not really helping legibility and maintainability, though. Perhaps if you want to make each part reasonably self-domumenting, use comments.

is_leap () {
    # Divisible by 4?
    (("$1" % 4 == 0)) || return
    # Not divisible by 100?
    (("$1" % 100 > 0)) && return
    # Divisible by 400?
    (("$1" % 400 == 0))
}

... Though the comments seem rather superfluous here.

edited Jan 27, 2019 at 7:22

answered Jan 26, 2019 at 20:58

tripleee

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