PHP 为什么不能将静态变量初始化为time()？

Question

我在将静态变量初始值设置为 time() 时收到错误消息。

错误信息是"语法错误，意外的'('，期待','或';'在[...][.. .] 第 7 行"

<?php 
define("DEBUG", true);
define("NEWLINE", "<br>");
...
function debugMsg($msg) {
    static $lastTime = time();
    if(DEBUG==true) echo date('Y-m-d H:i:s', time())." ".$msg." (".(time()-$lastTime)." seconds)".NEWLINE;
    $lastTime = time();
}
?>

debugMsg("XXX task completed");
//My expected output like below:
//2015-05-01 15:04:47 XXX task completed (2 seconds)

Answer 1

来自 PHP 文档 link

Static variables may be declared as seen in the examples above. Trying to assign values to these variables which are the result of expressions will cause a parse error.

您正在尝试为作为表达式结果的变量赋值

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