难道是保证安全执行的memcpy（0,0,0）？ [英] Is it guaranteed to be safe to perform memcpy(0,0,0)?

问题描述

我不是在C标准这么精通，所以请多多包涵。

I am not so well-versed in the C standard, so please bear with me.

我想知道，如果它是保证，由标准，即的memcpy（0,0,0）是安全的。

I would like to know if it is guaranteed, by the standard, that memcpy(0,0,0) is safe.

我能找到的唯一限制是，如果内存区域重叠，则行为是不确定...

The only restriction I could find is that if the memory regions overlap, then the behavior is undefined...

但我们认为内存区域重叠在这里？

But can we consider that the memory regions overlap here ?

推荐答案

我有C标准（ISO / IEC 9899：1999）的草稿版本，它有一些有趣的东西，说这一呼吁。对于初学者来说，它提到（教派; 7.21.1 / 2）关于的memcpy 是

I have a draft version of the C standard (ISO/IEC 9899:1999), and it has some fun things to say about that call. For starters, it mentions (§7.21.1/2) in regards to memcpy that

当一个参数声明为为size_t n指定一个数组的长度
   功能，n可以有到该函数的调用的值为零。除非明确说明
  否则在一个特定的功能在本节的描述中，指针参数
  这样的电话仍然应具有有效的值，如7.1.4 描述。在这样的呼叫，
  即找到一个字符函数没有发生，比较两个函数
   字符序列返回零，那副本复印件字符零功能
  字符。

Where an argument declared as size_t n specifies the length of the array for a function, n can have the value zero on a call to that function. Unless explicitly stated otherwise in the description of a particular function in this subclause, pointer arguments on such a call shall still have valid values, as described in 7.1.4. On such a call, a function that locates a character finds no occurrence, a function that compares two character sequences returns zero, and a function that copies characters copies zero characters.

在这里表示参考点这样：

The reference indicated here points to this:

如果一个参数传递给函数的值无效（如价值
  的功能的程序的地址空间外域，或指针外，
  或一个空指针或当相应的指针不可修改的存储
  参数不是常量限定）或类型（升级后）没有一个函数预期
  带有可变数量的参数，这种行为是未定义

If an argument to a function has an invalid value (such as a value outside the domain of the function, or a pointer outside the address space of the program, or a null pointer, or a pointer to non-modifiable storage when the corresponding parameter is not const-qualified) or a type (after promotion) not expected by a function with variable number of arguments, the behavior is undefined.

因此​​，它看起来是依照C规格，呼吁像

So it looks like according to the C spec, calling

memcpy(0, 0, 0)

在不确定的行为，因为空指针被认为是无效值。

results in undefined behavior, because null pointers are considered "invalid values."

这是说，我简直令人吃惊，如果你这样做的的memcpy 任何实际的实现爆发，由于大多数直观的实现我能想到的将在什么都不做所有，如果你说要复制零字节。

That said, I would be utterly astonished if any actual implementation of memcpy broke if you did this, since most of the intuitive implementations I can think of would do nothing at all if you said to copy zero bytes.

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