在PHP中，在传递对函数的引用时声明其变量类型是可选的吗？

Question

im new to PHP and i have come across two ways that people declare their functions parameters.

function fun(array &$array)
{
   return $stuff
}

OR

function fun($array)
{
   return $stuff
}

From what i read in the documentation the & passes the reference to the variable. but i thought that declaring the type array was unnecessary.

So, when i pass reference to a variable in a function is declaring its variable type optional?

Meaning is function fun(array &$array) AND function fun(&$array) equivalent/allowed?

Answer 1

To answer the question, Its optional because not all values are allowed. You can specify non scalar values (arrays, objects etc) but not ints strings floats etc

Your last line they are the same, with the exception that a catchable fatal error is thrown if its called like this

fun("1")
fun(false)

the way without the type hint would allow the function to be run and the functions logic would have to test that its argument is correct