的foreach（）：无效的参数传递

Question

我试图运行下面的代码，并得到一个错误“）提供的foreach（无效参数”：

<?php 
function xrange($start, $limit, $step = 1) { 
    if ($start < $limit) { 
     if ($step <= 0) { 
      throw new LogicException('Step must be +ve'); 
     } 

     for ($i = $start; $i <= $limit; $i += $step) { 
      yield $i; 
     } 
    } else { 
     if ($step >= 0) { 
      throw new LogicException('Step must be -ve'); 
     } 

     for ($i = $start; $i >= $limit; $i += $step) { 
      yield $i; 
     } 
    } 
} 

/* 
* Note that both range() and xrange() result in the same 
* output below. 
*/ 

echo 'Single digit odd numbers from range(): '; 
foreach (range(1, 9, 2) as $number) { 
    echo "$number "; 
} 
echo "\n"; 

echo 'Single digit odd numbers from xrange(): '; 
foreach (xrange(1, 9, 2) as $number) { 
    echo "$number "; 
} 
?> 

任何人都可以提出什么样的根本原因是作为即时通讯无法找出？

Answer 1

yield关键字只能用于5.5版本以后的PHP。或者，你可以修改你的功能：

function xrange($start, $limit, $step = 1) { 
    $num = array(); 
    if ($start < $limit) { 
     if ($step <= 0) { 
      throw new LogicException('Step must be +ve'); 
     } 

     for ($i = $start; $i <= $limit; $i += $step) { 
      $num[]= $i; 
     } 
    } else { 
     if ($step >= 0) { 
      throw new LogicException('Step must be -ve'); 
     } 

     for ($i = $start; $i >= $limit; $i += $step) { 
      $num[]= $i; 
     } 
    } 
    return $num; 
} 

，这将工作得很好。