使用malloc(0)和memcpy

我在某处读到：取消引用由大小为零的“新”分配返回的指针是UB.
在C中是否相同？
如果是,是以下代码UB？ (假设大小= 0)

a->object[index].data = malloc(size);  
 memcpy(a->object[index].data,bytes,size);

据我了解：不.只是想仔细检查一下.

解决方法

当您将0作为参数传递给malloc时,它将分配的内存释放到malloc返回的指针.

结果是实现定义的.

C11：7.22.3内存管理功能：

[…] If the size of the space requested is zero,the behavior is implementation-defined: either a null pointer is returned,or the behavior is as if the size were some nonzero value,except that the returned pointer shall not be used to access an object.

标准也说：

The free function causes the space pointed to by ptr to be deallocated,that is,made available for further allocation. If ptr is a null pointer,no action occurs.

因此,在实现定义行为的任​​何一种情况下,释放都不会调用未定义的行为.

现在转到问题的另一部分.

7.1.4库函数的使用：

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.

C11：7.24.1 p(2)：

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.