std::memmove

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

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

Parameters

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 std::memcpy

Where strict aliasing prohibits examining the same memory as values of two different types, std::memmove may be used to convert the values.

Example

#include <cstring>
#include <iostream>
 
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';
}

1234567890
1234456890

See also