关于PHP开发中的Fatal error: Call-time pass-by-reference has been removed in ... 错误
2013-08-29 09:38
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allow_call_time_pass_referenceboolean
在函数调用时参数被按照引用传递时是否发出警告。此方法已不被赞成并在 PHP/Zend 未来的版本中很可能不再支持。鼓励使用的方法是在函数定义中指定哪些参数应该用引用传递。鼓励大家尝试关闭此选项并确保脚本能够正常运行,以确保该脚本也能在未来的版本中运行(每次使用此特性都会收到一条警告)。
在函数调用时通过引用传递参数是不推荐的,因为它影响到了代码的整洁。如果函数的参数没有声明作为引用传递,函数可以通过未写入文档的方法修改其参数。要避免其副作用,最好仅在函数声明时指定那个参数需要通过引用传递。
参见引用的解释。
版本 | 说明 |
---|---|
5.4.0 | 从 PHP 中移除。 |
5.3.0 | Emits an E_DEPRECATEDlevel error. |
5.0.0 | Deprecated, and generates an E_COMPILE_WARNINGlevel error. |
PHP在升级到5.4版本的php可能会出现这种错误:
如果这样使用函数(或者类)的话,会产生一个 PHP Fatal error:
foo(&$var);
实际上,这样用法在php5.3中就会有提示,只是之前的仅仅会提示Deprecated而已。
// 正确写法 function myFunc(&$arg) { do something... } myFunc($var);//Call myFunc //错误写法 function myFunc($arg) { do something... } myFunc(&$arg);//Call myFunc
************************************************************************************************************************************************************************************
意思是调用时引用传参已经被移除了,就是不能通过function(&$a)这种方式传参调用函数. 解决方案: 查看你的php.ini配置文件,把其中的 allow_call_time_pass_reference参数调整为true,并重启服务器试试。 ====================================== 此外,以前的php代码在升级到5.4版本的php可能会出现这种错误: 当我们这样使用函数(或者类)的话,会产生一个error: foo(&$var); 实际上,这样用本来就是错的,只是之前的错误级别仅仅是Deprecated而已。 而正确的使用方法应该是在函数定义时: function foo(& $var) { //other code } 而在调用时直接传参就行了: foo($var);
引用:
http://www.php.net/manual/zh/ini.core.php
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