Make the function return 0 and store it in a variable

Question

Seems super trivial, but can't find a solution to this specific case on SO

A function may return a value of 0 OR another number, which I then want to store in a variable $s to calculate stuff. But can't find a clean way of doing it.

So for example:

function f() {
    $v = "0";
    return $v;
}

if($s = f() !== false) {
    echo $s;
    // ^ I want 0, but the value above is 1 (since it's true)
}

I tried returning it as a string

return "0" instead of a digit, but it doesn't work.

If I do this it will not evaluate to true so nothing will happen

if($s = f()) {
   // returns nothing
}

But when I var_dump(f()), it does show string '0' (length=1)

So I can do

if(var_dump(f()) == 0)

OR

if(f() == 0)

But is there not a cleaner way to do it, so the function may return 0 or another number and I can just capture it in a variable?

damn it.. thank you! Feel so dumb now. Can you post this as an answer, I ran into this a few times before and I think others will find this useful — Robert Sinclair, Apr 18 '19 at 03:09
@RobertSinclair this is what almost all php developer feels :D — NullPoiиteя, Apr 18 '19 at 03:15
amen brotha, i'll bookmark this. Have ran into this a few times when querying tables — Robert Sinclair, Apr 18 '19 at 03:16
I don't think it's specific to PHP. I think all languages would do that due to operator precedence (comparison is higher than assignment usually). `bool b = 3 == 5;` would almost always never compare `b = 3` to `5` unless you specifically told it to with parenthesis. — , Apr 18 '19 at 03:23

score 2 · Accepted Answer · answered Apr 18 '19 at 03:13

2

Add parentheses:

if (($s = f()) !== false) {

Otherwise you're computing the value f() !== false and assigning it to $s, instead of assigning f() to $s and then comparing to false.

answered Apr 18 '19 at 03:13

hobbs

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score 2 · Answer 2 · answered Apr 18 '19 at 03:20

First

if($s = f() !== false) {
echo $s;
// ^ I want 0, but the value above is 1 (since it's true)
}

Whats happening here is return value of f() is strictly compared to false, that is, "0" is compared to false. They are not strictly equal, hence f() !== false returns true, which is stored in $s as 1.

Next,

if($s = f()) {
// returns nothing
}

This doesnt enter the if block because $s contains "0" and it is loosely equal to false.

if(var_dump(f()) == 0) or if(f() == 0)

this works because it is loosely comparing "0" to 0 which is true. Hence it enters the block.

Remember, if condition does loose comparision or == and === does strict comparision. More about it here.

So, this should work in your case

if(($s = f()) !== 0)

if f() returns integers

Thanks man, great explanation, accepted the other answer since he posted working code first. But appreciate your time. — Robert Sinclair, Apr 18 '19 at 03:22

score 1 · Answer 3 · answered Apr 18 '19 at 03:10

1

Try some parenthesis:

if(($s = f()) !== false) {

To force $s to equal f(), not f() !== false

answered Apr 18 '19 at 03:10

Make the function return 0 and store it in a variable

3 Answers3