通过引用和三元运算符分配变量？ [英] Assigning variables by reference and ternary operator?

问题描述

为什么三元运算符不能通过引用进行分配？

Why ternary operator doesn't work with assignment by reference?

$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);

推荐答案

在这里

$target = ($result ? &$success : &$errors);

您的示例也有两个错字

http://php.net /manual/zh-CN/language.operators.comparison.php

注意：请注意，三元运算符是一个表达式，但它没有不是对变量求值，而是对表达式的结果求值。了解是否要通过引用返回变量很重要。语句返回$ var == 42？ $ a：$ b;因此，按引用返回功能中的命令将不起作用，并且在更高版本的PHP中会发出警告。

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. This is important to know if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a return-by-reference function will therefore not work and a warning is issued in later PHP versions.

idk如果以前曾起作用，但它不能不再了。如果您不想使用if语句，请尝试以下操作：

idk if this worked before, but it doesn't anymore. if you don't wanna use an if statement, then try this:

$result ? $target = &$success : $target = &$errors;

或在单独的行上...

or on separated lines ...

$result 
  ? $target = &$success 
  : $target = &$errors;

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