Re: copy smaller array into bigger array?
On Aug 15, 5:46 pm, "Fraser Ross" <z...@zzzzzz.com> wrote:
"Francesco"
Is copy_n really part of the STL?
Its in the latest draft standard. Some libraries might have
it within the TR1 namespace.
The following
page:http://www.cppreference.com/wiki/stl/algorithm/copy_n
States:
"copy_n
This function was part of the original SGI STL library, but
never has been a part of ISO C++."
Is it accurate and up to date?
Its out of date. Its quite well known that copy_if was
omitted by the standard.
And copy_n, apparently. But it has since been reinstated.
What do you thing about this version, which allows copying a
smaller array to the beginning of a larger one and uses
std::copy which is surely part of the ISO STL?
-------
template <typename T, size_t Nfrom, size_t Nto>
inline bool copy(T (&from)[Nfrom], T (&to)[Nto]) {
if (Nto < Nfrom) {
return false;
} else {
std::copy(from, &from[Nfrom], to);
The second argument contains undefined behavior. Use from +
Nfrom instead.
return true;
}
}
The OP has arrays of a single size. I don't think your
function has any use as it is.
It's certainly useful for the case the original poster
presented--copying a one dimensional array into a row of a two
dimensional array. For his case, in fact, something like:
assert( Nto == Nfrom ) ;
std::copy( from, from + nFrom, to ) ;
would be even more appropriate.
Nfrom and Nto are compile-time constants and you have a
run-time comparison of them.
Agreed. Some form of static assert would be more appropriate.
You have taken the address of a non-existing element
from[Nfrom] which is usually doable but isn't good. You can
write from + Nfrom.
It's a good habit to get into. Taking the address of one past
the end of a C style array will usually work, even if it is
formally undefined behavior. Taking the address of one past the
end of any other container in this manner, however, will
normally cause the program to crash, at least with a good
implementation of the library.
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