std::memmove
Defined in header <cstring>
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void* memmove( void* dest, const void* src, std::size_t count ); |
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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
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
Output:
1234567890 1234456890
See also
copies one buffer to another (function) | |
fills a buffer with a character (function) | |
copies a certain amount of wide characters between two, possibly overlapping, arrays (function) | |
(C++11) |
copies a range of elements to a new location (function template) |
copies a range of elements in backwards order (function template) | |
(C++11) |
checks if a type is trivially copyable (class template) |
C documentation for memmove
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