Exclamation mark in front of variable - clarification needed

问题

I've been working with PHP for quite a while now, but this was always a mystery to me, the correct use of the exclamation mark (negative sign) in front of variables.

What does !$var indicate? Is var false, empty, not set etc.?

Here are some examples that I need to learn...

Example 1:

$string = 'hello';
$hello = (!empty($string)) ? $string : '';

if (!$hello)
{
    die('Variable hello is empty');
}

Is this example valid? Would the if statement really work if $string was empty?

Example 2:

$int = 5;
$count = (!empty($int)) ? $int : 0;

// Note the positive check here
if ($count)
{
   die('Variable count was not empty');
}

Would this example be valid?

I never use any of the above examples, I limit these if ($var) to variables that have boolean values only. I just need to know if these examples are valid so I can broaden the use of the if ($var) statements. They look really clean.

Thanks.

回答1:

if(! $a) is the same as if($a == false). Also, one should take into account that type conversion takes place when using == operator.
For more details, have a look into "Loose comparisons with ==" section here. From there it follows, that for strings "0" and "" are equal to FALSE ( "0"==false is TRUE and ""==false is TRUE, too).

Regarding posted examples:
Example 1
It will work, but you should note, that both "0" and "" are 'empty' strings.

Example 2
It will work