php - 为什么@unset 给出解析错误？

Question

为什么在调用 unset 时不能使用 @ 运算符隐藏错误？以下导致解析错误:

@unset($myvar);

Answer 1

@ 运算符仅适用于表达式，unset 是一种语言构造，而不是函数。查看manual page了解更多信息:

Note: The @-operator works only on expressions. A simple rule of thumb is: if you can take the value of something, you can prepend the @ operator to it. For instance, you can prepend it to variables, function and include() calls, constants, and so forth. You cannot prepend it to function or class definitions, or conditional structures such as if and foreach, and so forth.