PHP函数中默认参数的的写法

函数可以定义 C++ 风格的标量参数默认值,如下所示: Example #3 在函数中使用默认参数



<?php function makecoffee($type = "cappuccino") { return "Making a cup of $type.\n"; } echo makecoffee(); echo makecoffee(null); echo makecoffee("espresso"); ?> 以上例程会输出: Making a cup of cappuccino. Making a cup of . Making a cup of espresso. PHP 还允许使用数组 array 和特殊类型 NULL 作为默认参数,例如: Example #4 使用非标量类型作为默认参数



<?php function makecoffee($types = array("cappuccino"), $coffeeMaker = NULL) { $device = is_null($coffeeMaker) ? "hands" : $coffeeMaker; return "Making a cup of ".join(", ", $types)." with $device.\n"; } echo makecoffee(); echo makecoffee(array("cappuccino", "lavazza"), "teapot"); ?> 默认值必须是常量表达式,不能是诸如变量,类成员,或者函数调用等。  注意当使用默认参数时,任何默认参数必须放在任何非默认参数的右侧;否则,函数将不会按照预期的情况工作。考虑下面的代码片断: Example #5 函数默认参数的不正确用法



<?php function makeyogurt($type = "acidophilus", $flavour) { return "Making a bowl of $type $flavour.\n"; } echo makeyogurt("raspberry");   // won't work as expected

?> 以上例程会输出: Warning: Missing argument 2 in call to makeyogurt() in /usr/local/etc/httpd/htdocs/phptest/functest.html on line 41 Making a bowl of raspberry . 现在,比较上面的例子和这个例子: Example #6 函数默认参数正确的用法



<?php function makeyogurt($flavour, $type = "acidophilus") { return "Making a bowl of $type $flavour.\n"; } echo makeyogurt("raspberry");   // works as expected

?> 以上例程会输出: Making a bowl of acidophilus raspberry. Note: 自 PHP 5 起,传引用的参数也可以有默认值。

 

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