about issues with objects in the mysqli extension library

About object issues in the Mysqli extension library
$mysqli _stmt= $mysqli->prepare ($sql);

The above prepare belong to the Mysqli_stmt class? Why directly with $mysqli->prepare? Two is all right.
That should be the truth.
$mysqli _stml=new mysqli_stmt ();
$stmt = $mysql _stmt->prepare ();

------Solution--------------------
Mysqli_stmt mysqli::p repare (String $query)
The definition in the manual can you find it?
It means that prepare belongs to the Mysqli class, the argument is a string type, and the return value is an instance of the Mysqli_stmt class.
------Solution--------------------
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Still don't understand. Directly with this sentence $stmt= $mysqli->prepare ($sql); is equivalent to creating an object $stmt? But the object should be created with new. I can't understand this place. Just got the wrong call.

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