std::memmove

Performs the following operations in order:

1. Implicitly creates objects at dest.
2. Copies count characters (as if of type unsigned char, the same below) from the object pointed to by src into a temporary array arr of count characters, where arr does not overlap the objects pointed to by dest and src.
3. Copies count characters from arr into the object pointed to by dest.

If dest or src is a null pointer or invalid pointer, the behavior is undefined.

[edit] Parameters

[edit] Return value

If there is a suitable created object, returns a pointer to it; otherwise returns dest.

[edit] Notes

Despite the specification says 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.

[edit] 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

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

[edit] See also