Streaming directly between streams

I was wondering if you can stream directly from one stream to another. e.g.

Code:

---

ifstream f("somefile", ios::in);
cout << f;

---

which obviously didn't work on my compiler (g++).

I know you can use rdbuf or getline or get.

I tried something called "tie" but that didn't work either:

Code:

---

cout.tie(&ifstream);

---

So how do you go about doing this?

Maybe this is something like what you want to do?

Code:

---

# include <fstream>
# include <iostream>
# include <iomanip>
# include <iterator>
# include <algorithm>
using namespace std;

int main()
{
    ifstream f("test.txt",ios::in);

    f >> noskipws;
    std::copy(istream_iterator<char>(f), istream_iterator<char>(), ostream_iterator<char>(cout, ""));

    return 0;
}

---

If you're going to do that you might as well use rdbuf.

using namespace std;

std::copy(...);


Heh, just leave out the using namespace std; it is a poor habit anyway.

The ifstream, the cout and the iterators need an std:: then.

True Daved, but my point still stands, leave out using namespace std; That is only there so that old code written prestandard can still compile.

I personally always use std::, but use of the using directive here is pretty harmless and not the point of the code at all.

Nothing wrong with using rdbuf(), you know ...

So, I take it that there is no way to stream directly from stream to stream using the << operator? Well, the copy method was interesting enough so thanks folks.

>> So, I take it that there is no way to stream directly from stream to stream using the << operator?
There is using rdbuf, but you already knew that.

Perhaps I should look at the specifications for this, but I was always curious why you had to use function calls to do IO on streams.

Why won't

Code:

---

//f is an fstream
cout << f;

---

implicitly understand that I just want to dump the entire thing to cout?

Why do you need the function calls, like:

Code:

---

//assuming ss is any type of stream
cout << ss.rdbuf();
//somestring would be a variable of type string
cout << somestring.c_str();

---

If anybody could throw me a few links explaining why this is so, I'd appreciate it.

it has to do with the fact << is actually an operator and the object cout has it overloaded accepting only a limited range of right-hand operands.

http://www.cplusplus.com/ref/iostrea...ratorltlt.html

one of them is a stream buffer

you could always overload the << operator yourself.

something like

Code:

---

ostream &operator<<(ostream &os, ifstream &f)
{
    os << f.rdbuf();
    return os;
}

---

but that doesn't really gain you much and could possibly cause an overload ambiguity (or maybe not, I'm just thinking aloud)

>> cout << somestring.c_str();
You don't need a function call there, there is an overload for the string class already:

Code:

---

cout << somestring;

---

Perhaps you forgot to #include <string>. On some compilers the string class works for the most part but fails with operator<< if you don't include the proper header.