Defined in header
void* memmove( void* dest, const void* src, std::size_t count );
Copies count characters from the object pointed to by src to the object pointed to by dest. Both objects are reinterpreted as arrays of unsigned char.
The objects may overlap: copying takes place as if the characters were copied to a temporary character array and then the characters were copied from the array to dest.
If either dest or src is an invalid or null pointer, the behavior is undefined, even if count is zero.
|dest||-||pointer to the memory location to copy to|
|src||-||pointer to the memory location to copy from|
|count||-||number of bytes to copy|
 Return value
std::memmove may be used to implicitly create objects in the destination buffer.
Despite being specified "as if" a temporary buffer is used, actual implementations of this function do not incur the overhead of double copying or extra memory. For small count, it may load up and write out registers; for larger blocks, a common approach (glibc and bsd libc) is to copy bytes forwards from the beginning of the buffer if the destination starts before the source, and backwards from the end otherwise, with a fall back to std::memcpy when there is no overlap at all.
Where strict aliasing prohibits examining the same memory as values of two different types,
std::memmove may be used to convert the values.
 See also
| copies one buffer to another |
| fills a buffer with a character |
| copies a certain amount of wide characters between two, possibly overlapping, arrays |
| copies a range of elements to a new location |
| copies a range of elements in backwards order |
| checks if a type is trivially copyable |
C documentation for memmove