NAME

std::memmove - std::memmove

Synopsis

Defined in header <cstring>
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.

If the objects are potentially-overlapping or not TriviallyCopyable, the behavior of
memmove is not specified and may be undefined.

Parameters

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

dest

Notes

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.

Example

// Run this code

#include <iostream>
#include <cstring>

int main()
{
char str[] = "1234567890";
std::cout << str << '\n';
std::memmove(str + 4, str + 3, 3); // copies from [4, 5, 6] to [5, 6, 7]
std::cout << str << '\n';
}

Output:

1234567890
1234456890