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关于摘引变量的疑惑

WBOY
Release： 2016-06-13 11:29:04
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关于引用变量的疑惑
$foo='Bob';
$bar=&$foo;
$bar="my name is $bar";
echo $bar.'
';
echo $foo;

这是手册上的一个例子，我有疑惑的就是输出$bar的结果我能理解，但是为何输出$foo的时候，他的值不是Bob，而是my name is Bob，为何其值会变，请高手解释，谢谢！

Related labels：
bar echo nbsp
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