PHP static variable testing, beginner PHP static variable error analysis

function myfunc()
{
static $int;
$int=0;
";
}
echo myfunc();
echo myfunc();
echo myfunc();
?>< ;/p>

The three values ​​in the book are 1, 2, and 3 respectively However, the actual result is that it cannot be run, and the syntax is wrong. The reason for the post-check error is that the sentence $int+1."
" should be written as ($int+1)."
". After changing it, the program no longer reports an error. But the values ​​are 1, 1, 1; in fact, this is not difficult to explain. Although $int keeps adding 1, the result is not assigned to $int again. Why does $int increase?

Modify the code to the following to make it correct:

function myfunc()
{
static $int=0; //php static variable definition
$int=$int+1;
echo $int."}
echo myfunc();
echo myfunc();
echo myfunc();
?>

Note that the static keyword must be together with the assignment (php static static variable modifier usage), if it is written in the book

staitc $int; $int=0;

Error, the result after running is also 1, 1, 1