std::memmove

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 the objects are not TriviallyCopyable, the behavior of memmove is not specified and may be undefined.

Parameters

Return value

dest.

Notes

Despite being specified "as if" a temporary buffer is used, actual implementations of this function do not incur the overhead or double copying or extra memory. 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 the more efficient std::memcpy when there is no overlap at all.

Example

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

1234567890
1234456890

See also