c - 是否保证执行 memcpy(0,0,0) 是安全的？

Question

我对C标准不是很精通，所以请多多包涵。

我想知道根据标准是否可以保证 memcpy(0,0,0) 是安全的。

我能找到的唯一限制是，如果内存区域重叠，则行为未定义...

但是我们可以认为内存区域在这里重叠吗？

Answer 1

我有一个 C 标准的草案版本 (ISO/IEC 9899:1999)，它有一些关于那个调用的有趣的事情要说。对于初学者，它提到 (§7.21.1/2) 关于 memcpy

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.

此处指示的引用指向:

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.

所以它看起来像根据 C 规范，调用

memcpy(0, 0, 0)

导致未定义的行为，因为空指针被认为是“无效值”。

就是说，如果您这样做的话，如果 memcpy 的任何实际实现失败，我会感到非常惊讶，因为如果您说要复制，我能想到的大多数直观实现根本不会做任何事情零字节。