validity of begin() on empty string or vector

[Old Wolf]

Is this undefined behaviour:

#include <string>
#include <vector>
#include <algorithm>

int main()
{
std::string s;
char buf[20];

std::copy(s.begin(), s.end(), buf);
}

Same question again for std::vector<char> instead of std::string.

Do I have to go s.empty() before every time I go s.begin(), in
case begin() is undefined?

 

[Dietmar Kuehl]

Is this undefined behaviour:
std::string s;
char buf[20];

std::copy(s.begin(), s.end(), buf);

No, why should it? It is an empty sequence. This applies to all
container types which can be empty as well. Of course, you cannot
dereference the the past the end iterator which is in this case
identical to 'begin()' but this is not done by 'std::copy()' or
other standard algorithms taking an input sequence.

 

[Kristo]

Is this undefined behaviour:

#include <string>
#include <vector>
#include <algorithm>

int main()
{
std::string s;
char buf[20];

std::copy(s.begin(), s.end(), buf);
}

It's only undefined if s.size() is greater than 20, in which case you'd
overflow the buffer. Also note that buf will not be usable as a C-style
string after the call to std::copy(). To use it as such you'll have to
append a null character yourself.

Kristo

 

[Clark S. Cox III]

Is this undefined behaviour:

#include <string>
#include <vector>
#include <algorithm>

int main()
{
std::string s;
char buf[20];

std::copy(s.begin(), s.end(), buf);
}

Same question again for std::vector<char> instead of std::string.

Do I have to go s.empty() before every time I go s.begin(), in
case begin() is undefined?

No. If the container is empty, then begin() and end() return iterators
to the same location. Which means that they are valid in the sense that
they are non-singular, they simply cannot be dereferenced.

There is nothing undefined about the code you posted (of course, it
doesn't really do anything, but that's beside the point).