Operator ! in if ternary

Question

Could someone explain to me the use of ! at the beginning of the code?

$number = 28;

$resultado = !($number % 2) ? "O número é par" : "O número é ímpar";
echo $resultado;

Answer 1

In this code the use of the negation operator is an error or at least abuse. Even if it produces the expected result does not make sense and only does not give error of execution because PHP is a weak typing language.

The negation operator takes a boolean value, therefore false or true and reverses its value.

The result of $number % 2 is a number and not a boolean so it is not correct to apply the negation operator to it. Works and in some cases can produce what you want, but is confusing.

In PHP any value 0 (or equivalent to 0) is always considered false and any other value is considered true, so the language makes an automatic coercion of type following this rule (documentation).

This is a case that even works because the result will always be 0 or 1 so coercion works well. So it would be more explicit what is the condition you want to be false or true:

$number % 2 == 0 ? "O número é par" : "O número é ímpar";

This way the code is very clear that you want it to be true when the result is 0, and we know that the rest of a division by 2 is always 0 when the number per pair, if 1 (therefore different from 0) is odd.

And if you really want to use the result as boolean you don’t have to do the negation:

$number % 2 ? "O número é ímpar" : "O número é par";

I put in the Github for future reference.

The cases that negation is necessary and useful are rare, unless it comes from a function that returns a boolean, but by an expression you can always build it differently to avoid the operator, even if it doesn’t always pay off. Readability to demonstrate intent is what comes first, make sure denial will help you better understand. This is not the case in the question.

Answer 2

In PHP, 1 and 0 are values truthy and falsy, they can be used as logical conditions, although their types are different from boolean.

In your example you have $number % 2. As the value of $number is 28, the rest of this division will be 0, that is, it is falsy, equivalent to false in that case of use.

Just in doing $resultado = false ? "O número é par" : "O número é ímpar", $resultado would be receiving O número é ímpar, which is not true, for 28 is par.

To solve this problem, you could reverse your ternary expression, or else reverse your result.

That is to say,

$resultado = $number % 2 ? "O número é ímpar" : "O número é par";

Or else

$resultado = !($number % 2) ? "O número é par" : "O número é ímpar";

In the latter case, the result of 28 % 2; 0, which corresponds to a value falsy, is denied, becoming true, soon $resultado receives O número é par, which is exactly what you want.