php object operator -> without instance?

Question

I have the following code in script I implement and it does work correctly, just want to understand:

 $variable1 = function1(); // function1 is a class method, it’s file included per require_once $ variable1 -> function2(); // function2 is a class method, it’s file included per require_once

Isn’t -> here an object operator? But there is no initialized instance save in $variable1.

Help appreciated

Naftali · Accepted Answer · 2012-04-04 13:49:28Z

That just means that function1() returns an object.

Hence you can use that object and it's functions.

Example:

class Test { function function2(){ echo "Hi"; } } function function1(){ return new Test; } //SO: $variable1 = function1(); $variable1->function2();

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