PHP static variable test, beginner PHP static variable error parsing

function MyFunc () { Static $int; $int = 0; echo $int +1. " "; } Echo MyFunc (); Echo MyFunc (); Echo MyFunc (); ?> Copy Code Three values in the book are three-in-one. However, the true result is not operational, syntax error, and post-check error is due to $int+1. " "The wording of this sentence should be written ($int + 1)." ", the program does not error after the change, but the value is 1,1,1; in fact, this is not difficult to explain, $int although the constant increase of 1, but the results have not been assigned to $int, talk about what $int will increase. The code is correct if you modify it to the following: function MyFunc () { static $int = 0; PHP static variable definition $int = $int +1; echo $int. " "; } Echo MyFunc (); Echo MyFunc (); Echo MyFunc (); ?> Copy Code Note that the static keyword must be combined with the assignment (the use of the PHP static variable modifier), if written in a book STAITC $int; $int = 0; Error, the result of the run is also 1,1,1

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