在php中取消设置变量 [英] unset variable in php

问题描述

我刚刚通过 php 手册阅读了有关未设置变量的信息.

php 手册说unset() 销毁指定的变量"

这个定义看起来很完美，直到我遇到了静态变量......如果函数内部的静态变量是 unset()，则 unset() 只会在函数其余部分的上下文中销毁该变量.随后的调用将恢复变量的先前值."

这个定义对我来说似乎不是一个好定义，至少，因为销毁变量"意味着变量不再与该内存位置相关联.

有没有其他人认为更好的定义是unset() 使变量超出当前范围"?我的意思是，与其指向生命周期，不如在这里使用词范围?

解决方案

让我们考虑一下功能:

function foo() {静态 $bar;$bar++;未设置($bar)；}富()；//静态 $bar 为 1富()；//静态 $bar 是 2

函数编译为:

<前>函数名:foo操作数:11编译变量:!0 = $barline # * op fetch ext 返回操作数---------------------------------------------------------------------------------2 0 > EXT_NOP4 1 EXT_STMT2 FETCH_W 静态 $0 'bar'3 ASSIGN_REF !0, $05 4 EXT_STMT5 POST_INC ~1 !06 免费 ~16 7 EXT_STMT8 UNSET_VAR !07 9 EXT_STMT10 > 返回空

一个变量实际上存在于对 foo() 的每个函数调用之外，并且在每次调用时，它都会被提取并且对它的引用被分配给 $bar.其实和这个很像:

function foo() {全球 $bar;$bar++;未设置($bar)；}

当您调用 unset() 时，您只会破坏您创建的引用，而不是基础值.

我没有确认，但我猜会发生这样的事情:

• 存储变量的底层表示(zval)，以便其引用计数为 1.
• 当 foo() 被调用时，符号 $bar 与这个 zval 相关联，它的引用计数增加到 2 并设置引用标志.立>
• 当 unset 被调用时，zval 的引用计数减少到 1，引用标志可能被清除并且符号 $bar 被删除.

请参阅引用计数基础知识.

I just read about unset variable through php manual.

The php manual says "unset() destroys the specified variables"

This def seems perfect until I came across static variable... "If a static variable is unset() inside of a function, unset() destroys the variable only in the context of the rest of a function. Following calls will restore the previous value of a variable. "

This definition doesn't seems a good one for me, at least, since "destroy the variable" implies that the variable is no longer associated with that memory location.

Does anyone else think a better definition would be "unset() makes the variable out of current scope"? I mean, rather than pointing towards lifetime, it's better to use word scope here?

解决方案

Let's consider the function:

function foo() {
    static $bar;
    $bar++;
    unset($bar);
}
foo(); //static $bar is 1
foo(); //static $bar is 2

The function compiles to:

function name:  foo
number of ops:  11
compiled vars:  !0 = $bar
line     # *  op                           fetch          ext  return  operands
---------------------------------------------------------------------------------
   2     0  >   EXT_NOP                                                  
   4     1      EXT_STMT                                                 
         2      FETCH_W                      static              $0      'bar'
         3      ASSIGN_REF                                               !0, $0
   5     4      EXT_STMT                                                 
         5      POST_INC                                         ~1      !0
         6      FREE                                                     ~1
   6     7      EXT_STMT                                                 
         8      UNSET_VAR                                                !0
   7     9      EXT_STMT                                                 
        10    > RETURN                                                   null

A variable actually exists outside each function call to foo() and, on each call, it's fetched and a reference to it is assigned to $bar. In fact, it's very similar to this:

function foo() {
    global $bar;
    $bar++;
    unset($bar);
}

When you call unset(), you're only destroying the reference you created, not the underlying value.

I didn't confirm, but what I'd guess that happens is this:

• The underlying representation of the variabe (the zval) is stored so that its reference count is 1.
• When foo() is called, the symbol $bar is associated with this zval, its reference count is increased to 2 and the reference flag is set.
• When unset is called, the zval has its reference count decreased to 1, the reference flag is probably cleared and the symbol $bar is removed.

See reference count basics.

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