为什么我不应该尝试使用“this” “删除此”之后的值

问题描述

在这一段中，使用 delete this 构造。列出了4个限制。

限制1到3看起来相当合理。但为什么是限制4那里，我不能检查它，与另一个指针，比较它与NULL，打印，铸造，做任何事情与它？

我的意思是这个是另一个指针。为什么我不能 reinterpret_cast 它到 int 或调用 printf c $ c>以输出其值？

解决方案

在删除指针后不能做任何事情的原因，或任何其他指针），是硬件可能（和一些旧机器）捕获试图加载无效的存储器地址到寄存器。即使它在所有现代硬件上都可以正常工作，标准说，你可以对无效指针（未初始化或删除）做的唯一的事情是分配给它（NULL，或者从另一个有效的指针）。




In this paragraph of C++ FAQ usage of delete this construct is discussed. 4 restrictions are listed.

Restrictions 1 to 3 look quite reasonable. But why is restriction 4 there that I "must not examine it, compare it with another pointer, compare it with NULL, print it, cast it, do anything with it"?

I mean this is yet another pointer. Why can't I reinterpret_cast it to an int or call printf() to output its value?

解决方案

The reason that you cannot do anything with a pointer after you delete it (this, or any other pointer), is that the hardware could (and some older machines did) trap trying to load an invalid memory address into a register. Even though it may be fine on all modern hardware, the standard says that the only thing that you can do to a invalid pointer (uninitialized or deleted), is to assign to it (either NULL, or from another valid pointer).

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