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std::memmove

From cppreference.com
< cpp‎ | string‎ | byte
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 not trivially-copyable (scalars, arrays, C-style structs), the behavior is undefined.

Contents

[edit] 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

[edit] Return value

dest

[edit] 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 memcpy when there is no overlap at all.

[edit] 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';
}

Output:

1234567890
1234456890

[edit] See also

copies one buffer to another
(function) [edit]
copies a range of elements to a new location
(function template) [edit]
copies a range of elements in backwards order
(function template) [edit]
checks if a type is trivially copyable
(class template) [edit]
C documentation for memmove