Print elements of vector

[Markus Dehmann]

I am wondering what is the "best" code to print out the elements of a
vector.
Assume they contain strings.

With "best", I mean shortest and most readable at the same time.

Of course, a for loop can be done. But I guess one can do it more
elegantly.
Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

But I don't like it: The line begins with "copy", and that's
confusing. It has a kind of misleading semantics.

Then better with for_each. How would that look like? Or is there a
better version?

I wonder why there is no to_string(const string& delimiter) function
in vector. Then, we could write:

cout << v.to_string("\n ") << endl;

Markus

[Matthew Del Buono]

"Markus Dehmann" <ma*******@gmx. de> wrote in message
news:c1******** *************** ***@posting.goo gle.com...

I am wondering what is the "best" code to print out the elements of a
vector.
Assume they contain strings.

With "best", I mean shortest and most readable at the same time.

Of course, a for loop can be done. But I guess one can do it more
elegantly.
Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

But I don't like it: The line begins with "copy", and that's
confusing. It has a kind of misleading semantics.

Then better with for_each. How would that look like? Or is there a
better version?

I wonder why there is no to_string(const string& delimiter) function
in vector. Then, we could write:

cout << v.to_string("\n ") << endl;

Markus

IMHO, use arrays directly. With direct access like that, you don't have to
worry about something like this. It would undoubtably be faster -- think
about it. On the inside, a vector is just an easy interface for a linked
list.

-- Matt

[John Harrison]

On Thu, 15 Jul 2004 12:36:41 -0400, Matthew Del Buono <Pl*******@ToMe .com>
wrote:

IMHO, use arrays directly. With direct access like that, you don't have
to
worry about something like this. It would undoubtably be faster -- think
about it. On the inside, a vector is just an easy interface for a linked
list.

A vector is an easy interface to an array. It has no similarly to a linked
list. And don't knock easy interfaces.

And both of the OP suggestions std::copy and std::for_each are applicable
to arrays as well, so maybe you do have something to worry about.

john

[Ali Cehreli]

On Thu, 15 Jul 2004 09:19:55 -0700, Markus Dehmann wrote:

Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

But I don't like it: The line begins with "copy", and that's confusing.
It has a kind of misleading semantics.

Here is a to_string that puts the delimiter between the objects, not
after the last one:

#include <sstream>
#include <string>

using namespace std;

template <class Collection>
string to_string(Colle ction const & collection,
char const * delimiter)
{
typedef typename Collection::con st_iterator Iter;

Iter beg = collection.begi n();
Iter const end = collection.end( );

if (beg == end)
{
return string();
}

std::ostringstr eam os;

os << *beg;
++beg;

for ( ; beg != end; ++beg)
{
os << delimiter << *beg;
}

return os.str();
}

template<class Collection>
string to_string(Colle ction const & collection,
string const & delimiter)
{
return to_string(colle ction, delimiter.c_str ());
}

#include <vector>
#include <list>
#include <iostream>

template <class Type, size_t N>
size_t arraySize(Type const (&)[N])
{
return N;
}

void string_test()
{
string const array[] = { "Hello", "cool", "world" };
vector<string> words(array, array + arraySize(array ));
cout << to_string(words , string(" ")) << "!\n";
}

void int_test()
{
int const array[] = { 1, 7, 42, 100 };
vector<int> numbers(array, array + arraySize(array ));
cout << to_string(numbe rs, " + ") << '\n';
}

void empty_test()
{
list<double> doubles;
cout << "Empty: " << to_string(doubl es, " - ") << '\n';
}

int main()
{
string_test();
int_test();
empty_test();
}

Ali

[Jeff]

ma*******@gmx.d e (Markus Dehmann) wrote in message news:<c1******* *************** ****@posting.go ogle.com>...

I am wondering what is the "best" code to print out the elements of a
vector.
Assume they contain strings.

With "best", I mean shortest and most readable at the same time.

Of course, a for loop can be done. But I guess one can do it more
elegantly.
Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

This is as good as it gets. There ain't nothin better.
But I don't like it: The line begins with "copy", and that's
confusing. It has a kind of misleading semantics.

Then better with for_each. How would that look like? Or is there a
better version?

I wonder why there is no to_string(const string& delimiter) function
in vector. Then, we could write:

cout << v.to_string("\n ") << endl;

This method would be intrinsically less efficient. It would be
dynamically filling and resizing a string, then pumping it to cout in
one shot. A slow down would be noticeable if the vector were very
large.

Whereas even with a very large array, the copy() approach would be
efficient with less memory overhead.

[David Hilsee]

"Markus Dehmann" <ma*******@gmx. de> wrote in message
news:c1******** *************** ***@posting.goo gle.com...

I am wondering what is the "best" code to print out the elements of a
vector.
Assume they contain strings.

With "best", I mean shortest and most readable at the same time.

Of course, a for loop can be done. But I guess one can do it more
elegantly.
Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

But I don't like it: The line begins with "copy", and that's
confusing. It has a kind of misleading semantics.

In the end, the for loop is probably the most readable, and it's only
slightly more verbose. But beauty is in the eye of the beholder, so I'm
sure others find the ostream_iterato r version readable. If you want a
one-line method for displaying the contents of a container that avoids
"misleading semantics", then you could write a trivial function.

template <class T>
void display_contain er( const T& v ) {
// display elements using method of choice
std::copy(v.beg in(), v.end(), std::ostream_it erator<typename
T::value_type>( std::cout, "\n"));
}

int main() {
std::vector<int > v;
// ...
display_contain er(v);
}

--
David Hilsee

[Siemel Naran]

"Jeff" <su********@gma il.com> wrote in message

ma*******@gmx.d e (Markus Dehmann) wrote in message

cout << v.to_string("\n ") << endl;

This method would be intrinsically less efficient. It would be
dynamically filling and resizing a string, then pumping it to cout in
one shot. A slow down would be noticeable if the vector were very
large.

Whereas even with a very large array, the copy() approach would be
efficient with less memory overhead.

Not necessarily. If v.to_string(v) or to_string(v) returns a proxy object,
then we can overload ostream's operator<< to print the vector elements to
the stream directly. The proxy object would store a pointer to the vector
along with the delimeter character. It's syntactic sugar for the same old
loop or std::copy thing.

[Marc]

Markus Dehmann wrote:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

Then better with for_each. How would that look like?

That is an interesting question. I don't know any way to do this without
writing your own class:

struct print {
ostream& out;
print(ostream& os): out(os) {}
template <typename T>
void operator()(T const& x){ out << x; }
};

(I removed the delimiter to simplify the example)

It should have been possible to create the object print(cout) using only
the functional capabilities of c++, but bind1st accepts only const
arguments so it cannot be used to bind cout as the first argument of <<
or ostream_iterato r<string>(cout ) as the first argument of =.

bind1st(mem_fun _ref<ostream&,o stream,T>(&ostr eam::operator<< ),cout)
would be nice if it worked, wouldn't it? (no working around the fact
that you have to explicit the template arguments for an overloaded
function, it seems)

Or (where typ stands for ostream_iterato r<T>):
bind1st(mem_fun _ref<typ&,typ,c onst T&>(&typ::opera tor=),typ(cout) )

I hope some day something similar will be possible... (bind1st is not
the only thing that prevents the second exemple from working as is, also
bind1st_ref and allowing reference to reference would be enough for the
first one)

[Jeff Flinn]

Markus,

"Markus Dehmann" <ma*******@gmx. de> wrote in message
news:c1******** *************** ***@posting.goo gle.com...

I am wondering what is the "best" code to print out the elements of a
vector.
Assume they contain strings.

With "best", I mean shortest and most readable at the same time.

Of course, a for loop can be done. But I guess one can do it more
elegantly.
Something like this would be possible:

copy(v.begin(), v.end(), ostream_iterato r<string>(cout , "\n"));

But I don't like it: The line begins with "copy", and that's
confusing. It has a kind of misleading semantics.
Agreed, and forces one to play games if a different terminating delimiter is
desired.
Then better with for_each. How would that look like? Or is there a
better version?

I wonder why there is no to_string(const string& delimiter) function
in vector. Then, we could write:

cout << v.to_string("\n ") << endl;
How about something like:

std::cout << range_stream( v.begin(), v.end(), "{ ", ", " " }\n" );
With the following definition of range_stream:

#if !defined(RangeS treamerHeaderIn cluded)
#define RangeStreamerHe aderIncluded

template< class tItr, class tDBeg, class tDInner, class tDEnd > class
range_streamer
{
tItr mBeg;
tItr mEnd;
tDBeg mDBeg;
tDInner mDInner;
tDEnd mDEnd;
public:
range_streamer( tItr aBeg, tItr aEnd, tDBeg aDBeg, tDInner aDInner, tDEnd
aDEnd )
: mBeg ( aBeg )
, mEnd ( aEnd )
, mDBeg ( aDBeg )
, mDInner( aDInner )
, mDEnd ( aDEnd )
{}

template< class tStream >
void OutputTo( tStream& s )const
{
if( mBeg != mEnd )
{
tItr lItr = mBeg;

s << mDBeg << *lItr;

for( ++lItr ; lItr != mEnd ; ++lItr )
{
s << mDInner << *lItr;
}
s << mDEnd;
}
}
};

template< class tItr, class tDBeg, class tDInner, class tDEnd >
inline range_streamer< tItr,tDBeg,tDIn ner,tDEnd>
range_stream( tItr aBeg, tItr aEnd, tDBeg aDBeg, tDInner aDInner, tDEnd
aDEnd )
{
return range_streamer< tItr,tDBeg,tDIn ner,tDEnd>( aBeg, aEnd, aDBeg,
aDInner, aDEnd );
}

template< class tStream, class tItr, class tDBeg, class tDInner, class tDEnd

inline tStream& operator<<( tStream& s, const
range_streamer< tItr,tDBeg,tDIn ner,tDEnd>& aRS )
{
aRS.OutputTo(s) ;

return s;
}

#endif //RangeStreamerHe aderIncluded