PHP三元运算符不解析类属性内部吗? [英] PHP ternary operator doesn't parse inside class property?

问题描述

我试图弄清楚为什么类内部的三元运算符不会解析.我认为，举个例子是最清楚的方法.

I'm trying to figure out why a ternary operator inside of a class will not parse. I think an example is the clearest way to show this.

这很好:

$a = array(
    'a' => 'foo',
    'b' => 1 ? 'true' : 'false',
    'c' => 'baz',
);
print_r($a);

/* Array
   (
       [a] => foo
       [b] => true
       [c] => baz
    )
 */

但这甚至无法解析:

<?php

class Junk {
    private static $a = array(
        'a' => 'foo',
        'b' => 1 ? 'true' : 'false',
        'c' => 'baz',
    );

    public static function printA() {
        print_r(self::$a);
    }
}

Junk::printA();

我收到以下消息:

PHP Parse error:  syntax error, unexpected '?', expecting ')' in junk.php on line 6

为记录起见，它在数组声明之外也不起作用:

For the record, it doesn't work outside of an array declaration either:

private static $a = 1 ? 'true' : 'false';

给出相同的错误消息.

为什么这行不通?解析引擎中是否只是一些奇怪的错误?我完全困惑. 三元运算符手册指出:运算符是一个表达式，应始终在数组赋值的右侧使用.如果有任何区别，我使用的是PHP 5.4.28.

Why doesn't this work? Is it just some weird bug in the parsing engine? I am completely baffled. The manual on ternary operators states that the operator is an expression, which should always work on the right-hand side of an array assignment. I'm on PHP 5.4.28, if that makes any difference.

推荐答案

类属性必须是常量表达式.

A class property must be a constant expression.

此声明可能包含初始化，但这 初始化必须是一个常数值-也就是说，它必须能够 在编译时进行评估，并且不得依赖于运行时 信息以便进行评估.

This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.

http://www.php.net/manual/zh/language.oop5.properties.php

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