[英]Assigning variables by reference and ternary operator?
Why ternary operator doesn't work with assignment by reference? 为什么三元运算符不能通过引用分配?
$obj = new stdClass(); // Object to add
$result = true; // Op result
$success = array(); // Destination array for success
$errors = array(); // Destination array for errors
// Working
$target = &$success;
if(!$result) $target = &errors;
array_push($target, $obj);
// Not working
$target = $result ? &$success : &$errors;
array_push($target, $obj);
Here you go 干得好
$target = ($result ? &$success : &$errors);
Also your example has two typos 你的例子也有两个拼写错误
http://php.net/manual/en/language.operators.comparison.php http://php.net/manual/en/language.operators.comparison.php
Note: Please note that the ternary operator is an expression, and that it doesn't evaluate to a variable, but to the result of an expression.
注意:请注意,三元运算符是一个表达式,它不会计算变量,而是表达式的结果。 This is important to know if you want to return a variable by reference.
知道是否要通过引用返回变量很重要。 The statement return $var == 42 ?
声明返回$ var == 42? $a : $b;
$ a:$ b; in a return-by-reference function will therefore not work and a warning is issued in later PHP versions.
因此,在返回引用函数中将不起作用,并且在以后的PHP版本中发出警告。
idk if this worked before, but it doesn't anymore. idk如果之前有效,但它不再存在了。 if you don't wanna use an if statement, then try this:
如果你不想使用if语句,那么试试这个:
$result ? $target = &$success : $target = &$errors;
or on separated lines ... 或分开的线......
$result
? $target = &$success
: $target = &$errors;
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