Re: cout << vector<string>
"Kai-Uwe Bux" <jkherciueh@gmx.net> ha scritto nel messaggio
news:49144af2$0$17068$6e1ede2f@read.cnntp.org...
Juha Nieminen wrote:
Jeff Schwab wrote:
template<typename T>
std::ostream& operator<<(std::ostream& out, std::vector<T> const& v) {
if (!v.empty()) {
typedef std::ostream_iterator<T> out_iter;
copy(v.begin(), v.end() - 1, out_iter( out, " " ));
out << v.back();
}
return out;
}
Is there some advantage of that code over a shorter and simpler:
template<typename T>
std::ostream& operator<<(std::ostream& out, std::vector<T> const& v) {
for(std::size_t i = 0; i < v.size()-1; ++i)
Nit: std::size_t is not guaranteed to be vector::size_type.
so could be
std::size_t != vector::size_type
and what could happen if in the definition
size_t sii=big;
int a[sii];
vector<int> b(a, a+sii);
out << v[i] << " ";
out << v.back();
I think, v.back() has undefined behavior if the v is empty.
return out;
}
Best
Kai-Uwe Bux
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