std::memmove (3) - Linux Manuals

std::memmove: std::memmove

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 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

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.

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