"Array item" and constant "define" shows only 0 and not the whole number

Question

I declared an array item "0859" but when I display the array item, it shows only the 0 and not the whole number "0859".

In constant define I set the value to "01238" the result is the same with my array, it shows only the value 0 and not the whole number 01238. I would like to know further about the reason why array item and define that starts with 0 in there item and values shows only the 0 and not the whole number

Array

<?php
  $num = [0859];
  echo $num;

  output: 0

  Why not the the output is "0859"
?>

Constant Define

<?php

  define(myNum, 01238);
  echo myNum;

  output: 0

  Why not the the output is "01238"

?>

I'm using Wamp PHP version 5.5.12

Please enlighten me why I get the output only 0 and not the whole number

Answer 1

如果你正在运行PHP7，你应该看一下this的答案。

如链接答案所述：

  

这来自对整数，特别是八进制，   在PHP7中处理（与PHP5相反）。

     

八进制文字无效

     

以前，包含无效数字的八进制文字是   默默地截断（0128被视为012）。现在，一个无效的八进制   literal将导致解析错误。

     

将它们用作字符串或实际整数

     

$ a = array（1,8,9,12）; //整数

     

$ a = array（“00001”，“00008”，   “00009”，“00012”）; //字符串

您可以在此处找到有关整数/八进制的更多信息：http://php.net/manual/en/language.types.integer.php

Answer 2

完全归功于Qirel及其答案 - https://stackoverflow.com/a/40736053/2288334

这可能是由于PHP如何处理以前默认解析为八进制文字的内容。基本上，PHP7认为你试图将八进制作为数组的唯一值传递，但它是无效的。

供参考： http://php.net/manual/en/migration70.incompatible.php#migration70.incompatible.integers.invalid-octals http://php.net/manual/en/language.types.integer.php