03-22-2008
How to return void function pointer
Hello all
im trying to build function that will return void function pointer
what is mean is ( not working )
the main function
PHP Code:
void * myClass::getFunction(int type){
if(type==1)
return &myClass::Test1;
if(type==2)
return &myClass::Test2;
}
void myClass::Test1(){
printf("nada1");
}
void myClass::Test2(){
printf("nada2");
}
what im doing wrong here ?
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LEARN ABOUT PHP
uopz_compose
UOPZ_COMPOSE(3) 1 UOPZ_COMPOSE(3)
uopz_compose - Compose a class
SYNOPSIS
void uopz_compose (string $name, array $classes, [array $methods], [array $properties], [int $flags])
DESCRIPTION
Creates a new class of the given name that implements, extends, or uses all of the provided classes
PARAMETERS
o $name
- A legal class name
o $classes
- An array of class, interface and trait names
o $methods
- An associative array of methods, values are either closures or [modifiers => closure]
o $properties
- An associative array of properties, keys are names, values are modifiers
o $flags
- Entry type, by default ZEND_ACC_CLASS
RETURN VALUES
EXAMPLES
Example #1
uopz_compose(3) example
<?php
class myClass {}
trait myTrait {}
interface myInterface {}
uopz_compose(
Composed::class, [
myClass::class,
myTrait::class,
myInterface::class
], [
"__construct" => function() {
/* ... */
}
]);
var_dump(
class_uses(Composed::class),
class_parents(Composed::class),
class_implements(Composed::class));
?>
The above example will output something similar to:
array(1) {
["myTrait"]=>
string(7) "myTrait"
}
array(1) {
["myClass"]=>
string(7) "myClass"
}
array(1) {
["myInterface"]=>
string(11) "myInterface"
}
PHP Documentation Group UOPZ_COMPOSE(3)