Consider this code:
$x = 1.4;
$i1 = 0.5;
$i2 = 0.4;
echo ($i1 + $i2 = $x); // Outputs 1.9
Why is this? I've tried searching for this kind of variable setup without results. Is the variable $i2
being ignored? Why use this over echo ($x + $i1);
? It outputs the same result.
The point is that it does two things in one statement.
It is shorthand for:
$i2 = $x;
echo ($i1 + $i2);
The assignment happens inline, saving the separate line. Not ideal style, but often used in if()
, while()
and other control statements.
that would be $i1 + the assignment.
The assignment evaluates to $x ($i2 = $x )
the end result is echo 0.5 + 1.4.
Even php has operator priorities http://php.net/manual/en/language.operators.precedence.php .
=
在+
之前处理,这意味着发生了这种情况:
echo ($i1 + ($i2 = $x));
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.