memcpy zero bytes into const variable - undefined behavior?

Question

In C and C++, is it undefined behavior to memcpy into a const variable when the number of bytes to be copied is zero?

int x = 0;
const int foo = 0;
memcpy( (void *)&foo, &x, 0 );

This question is not purely theoretical. I have a scenario in which memcpy is called and if the destination pointer points to const memory, then the size argument is guaranteed to be zero. So I'm wondering whether I need to handle it as a special case.

Answer 1

c c17

The older question Is it guaranteed to be safe to perform memcpy(0,0,0)? points out 7.1.4p1:

Each of the following statements applies unless explicitly stated otherwise in the detailed descriptions that follow: 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.

The prototype for memcpy is

void *memcpy(void * restrict s1, const void * restrict s2, size_t n);

where the first parameter is not const-qualified, and &foo points to non-modifiable storage. So this code is UB unless the description of memcpy explicitly states otherwise, which it does not. It merely says:

The memcpy function copies n characters from the object pointed to by s2 into the object pointed to by s1.

This implies that memcpy with a count of 0 does not copy any characters (which is also confirmed by 7.24.1p2 "copies zero characters", thanks Lundin), but it does not exempt you from the requirement to pass valid arguments.