This works:
stringstream temp;
temp << i;
result_stream << transform(temp.str());
(transform is a function that takes a string and returns a string; i is an int). However, my attempt to let C++11 create a temporary object without a name didn't work:
result_stream << transform((stringstream() << i).str());
I thought it would work, since the second << should just return the first argument and I'd be able to use str() on that. But I get this error:
error: 'class std::basic_ostream<char>' has no member named 'str'
I'm using g++ 4.8.1 (MinGW-W64).
Is there a way to accomplish this (i.e. write code like this using an unnamed temporary)? (The above code is a bit simplified, and the actual code involves using << on arguments other than int.)
This doesn't work because the second << is std::ostream &operator<<(std::ostream &, int); and so the return type is ostream& which has no member str().
You would have to write:
result_stream << transform( static_cast<stringstream &>(stringstream() << i).str() );
Update (2019): According to LWG 1203 the standard may be changed in future (and one major implementation already has) so that this code no longer works, and a simpler code works instead. See this question for detail.
In the interim period, apparently the following works on both old and new:
result_stream << transform( static_cast<stringstream &>(stringstream().flush() << i).str() );
// ^^^^^^^^
This should not be a performance penalty since flushing an empty stream has no effect...
operator<<() returns a reference to the base class std::ostream contained within the std::stringstream. The base class doesn't contain the str() method. You can cast it back down to a std::stringstream&:
result_stream << transform(static_cast<std::stringstream&>(std::stringstream() << i).str());
The result of the << operator on the temporary stringstream is an ostream. There is no str() method on an ostream.
Use to_string instead:
using std::to_string;
result_stream << transform(to_string(i));
You can define a helper to_string to handle objects not covered by std::to_string.
template <typename T>
std::string to_string (const T &t) {
std::ostringstream oss;
oss << t;
return oss.str();
}
For example, if you had a class Foo that understood redirection to an ostream, and f was an instance of Foo, then you could do:
result_stream << transform(to_string(f));
Try it online!
If you actually want to use a lot of redirection to build up a string before transforming, you could create a helper object for that as well.
struct to_string_stream {
std::ostringstream oss;
template <typename T>
auto & operator << (const T &t) { oss << t; return *this; }
operator std::string () const { return oss.str(); }
void clear () { oss.string(std::string()); }
};
Then, you could do something like:
to_string_stream tss;
result_stream << transform(tss << i << ':' << f);
Try it online!
Tried and failed to do this for C++11 (in 2009):
http://cplusplus.github.io/LWG/lwg-active.html#1203
libc++ went outlaw and implemented it anyway.
It is up for reconsideration, but can not possibly be standardized prior to 2017 (standardization is a glacial process).
Related
I was working on a function which looks like so:
inline std::ostream& mvUp(int n,std::ostream& ss){
char u[8];
sprintf(u,"\033[%dA",n);
ss<<u;
return ss;
}
and using it like so:
std::cout<<mvUp(1);
however it shows error:
std::cout<<mvUp(1);
| ^_____too few args in function call
^_______________no operator "<<" match these operands
I also tried: std::cout<<mvUp(1,std::cout); but still not working.
std::cout<<mvUp(1);
^_______________no operator "<<" match these operands
now when I try making it template,
template <int n>
inline std::ostream& mvUp(std::ostream& ss){
char u[8];
sprintf(u,"\033[%dA",n);
ss<<u;
return ss;
}
and use it: std::cout<<mvUp<1>, this works totally fine but the problem with this is that templates take const args.
Not able to figure out where am I getting wrong. Also how is it working in templates when I am not passing any args?
Modern C++ code uses std::string, and other classes. This makes implementing this kind of an overload trivial.
#include <string>
#include <sstream>
inline std::string mvUp(int n) {
std::ostringstream o;
o << "\033[" << n << "A";
return o.str();
}
Then, everything will work automatically:
std::cout<<mvUp(1);
Your mvUp returns std::string, and the existing << overload takes care of the rest.
The fact that std::ostream& mvUp(std::ostream& ss); std::cout << mvUp<1> works is not a peculiarity of C++ function call syntax.
std::cout has its operator<< overloaded to accept single-parameter functions like this one, and call them by passing itself as the first argument.
Given std::ostream& mvUp(int n,std::ostream& ss);, std::cout<<mvUp(1) doesn't work because your function doesn't have one parameter. And std::cout << mvUp(1,std::cout); doesn't work because your function returns std::ostream&, which can't be printed.
The generic solution is to make a class with overloaded operator<<. But, as the other answer suggests, here you can just make a function that returns std::string, and print its return value.
You are trying to create a custom ostream manipulator, similar to std::setw() and other manipulators from the <iomanip> library. But what you have written is not the correct way to implement that. Your manipulator needs to return a type that holds the information you want to manipulate the stream with, and then you need to overload operator<< to stream out that type. That will give you access to the ostream which you can then manipulate as needed.
Try something more like this:
struct mvUpS
{
int n;
};
mvUpS mvUp(int n) {
return mvUpS{n};
}
std::ostream& operator<<(std::ostream& ss, const mvUpS &s) {
return ss << "\033[" << s.n << "A";
}
Now std::cout << mvUp(1); will work as expected.
Demo
So I got stuck on a simple print-function today and I really don't know how to fix this problem. Basically I want to pass my Strings to a function in a std::cout-style like this:
foo(std::ostringstream() << "Hello" << " World!");
And from what I've read it should be possible by a function along the lines of
void foo(std::ostringstream& in)
but when implementing I get a somewhat strange behavior:
#include <iostream>
#include <sstream>
void foo(std::ostringstream& in)
{
std::cout << in.str();
}
int main()
{
std::ostringstream working;
working << "Hello" << " World!";
foo(working); // OK
std::ostringstream notWorking;
notWorking << "Hello";
foo(notWorking<<" World!"); // Not OK?!
return 0;
}
While the first call of foo seems fine and works like expected, the second one refuses to compile even though they should (from my perspective) be technically the same thing.
Error:
error C2664: 'void foo(std::ostringstream &)': cannot convert parameter 1 from 'std::basic_ostream<char,std::char_traits<char>>' to 'std::ostringstream &'
I'm using MS Visual Studio 2013 Express on Win7 x64
The overloads of shifting operators for IO operations take the base streams by reference. That is:
std::ostream& operator<<( std::ostream& os , foo myfo )
{
os << myfoo;
return os;
}
So use std::ostream instead of std::ostringstream as function parameter:
void f( std::ostream& stream );
It's simply because the second expression
notWorking<<" World!"
returns a std::ostream& not a std::ostringsream&.
The reason for your error is that you’re trying to implicitly downcast an ostream into an ostringstream because the result of the << operator is always ostream& regardless of the concrete string type. Changing foo() to take an ostream& ought to solve this, unless you need to rely specifically on ostringstream features in the implementation of foo, in which case you can dynamic_cast.
Also, by constructing a temporary ostringstream in your foo(std::ostringstream() << "Hello" << " World!"); example, you obtain an rvalue. foo() expects an lvalue reference to the stream. You might be able to get away with changing foo() to take an rvalue reference: std::ostream&&.
It makes sense to implement << for QString like:
std::ostream& operator <<(std::ostream &stream,const QString &str)
{
stream << str.toAscii().constData(); //or: stream << str.toStdString(); //??
return stream;
}
instead of writing
stream << str.toAscii().constData();
every time in the code.
However, since it is not in standard Qt library, I'm assuming there is any particular reason not to do so. What are the risks/inconvenience of overloading << as specified above?
If the << operator is included in the Qt library every client of the library will have to use the exact same implementation. But due to the nature of QString it is far from obvious this is what these clients want. Some people writing software interacting with legacy file in western europe may want to use Latin1() characters, US people may go with Ascii() and more modern software may want to use Utf8().
Having a single implementation in the library would restrict unacceptably what can be done with the whole library.
It's not necessary to implement such thing, as long as there exists a convenient solution like this one, involving QTextStream
QString s;
QTextStream out(&s);
out << "Text 1";
out << "Text 2";
out << "And so on....";
QTextStream is quite powerfull...
The accepted answer points out some valid reasons for why there is no operator<< function for QString.
One can easily overcome those reasons by providing some convenience functions and maintaining some state in an application specific namespace.
#include <iostream>
#include <QString>
namespace MyApp
{
typedef char const* (*QStringInsertFunction)(QString const& s);
char const* use_toAscii(QString const& s)
{
return s.toAscii().constData();
}
char const* use_toUtf8(QString const& s)
{
return s.toUtf8().constData();
}
char const* use_toLatin1(QString const& s)
{
return s.toLatin1().constData();
}
// Default function to use to insert a QString.
QStringInsertFunction insertFunction = use_toAscii;
std::ostream& operator<<(std::ostream& out, QStringInsertFunction fun)
{
insertFunction = fun;
return out;
}
std::ostream& operator<<(std::ostream& out, QString const& s)
{
return out << insertFunction(s);
}
};
int main()
{
using namespace MyApp;
QString testQ("test-string");
std::cout << use_toAscii << testQ << std::endl;
std::cout << use_toUtf8 << testQ << std::endl;
std::cout << use_toLatin1 << testQ << std::endl;
return 0;
}
Output:
test-string
test-string
test-string
I don't think there is any particular reason for excluding (nor including) this in the Qt library. Only problem that could possibly appear here is a possibility that std::ostream object could modify the contents of the parameter passed to std::ostream::operator<< function.
However, in the reference it is clearly stated that this function will modify the parameter if string buffer is passed - there is nothing about the other types, so I guess (and the common-sense is telling me) that operator<< will not modify char* parameter. Also, on this page there is nothing about modifying the passed object.
Last thing: instead of using QString::toAscii().constData(), you could use QString::toStdString() or qPrintable(const QString&) macro.
I don't know why this is erroring, but I'm just trying to add something "akin" to endl so that I can throw what's in an ostringstream to our debugger. I have the following:
class debug_stream_info
{
public:
debug_stream_info(int errorLine, char *errorFile, int level)
:m_errorLine(errorLine), m_errorFile(errorFile), m_logLevel(level)
{
}
friend std::basic_ostringstream<char>& operator<<(std::basic_ostringstream<char>& os, debug_stream_info& debug_info);
private:
int m_errorLine;
std::string m_errorFile;
int m_logLevel;
};
std::basic_ostringstream<char>& operator<<(std::basic_ostringstream<char>& os, debug_stream_info& debug_info)
{
// Write the stream's contents to cpu_debug
// Deleted custom logging function. No errors here though
// Clear the stream for re-use.
os.str("");
os.seekp(0);
return os;
}
int main(int argc, char** argv)
{
std::ostringstream myout;
myout << "hey there" << " and some more " << "Numbers!!: " << 435 << 54.2 << " that's good for numbers" << debug_stream_info(__LINE__, __FILE__, LOG_LEVEL);
return 0;
}
The error I'm getting is: error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'debug_stream_info' (or there is no acceptable conversion) for the line in main. This is on VS2008.
I'm including sstream, iostream, etc, and have the namespaces set up right. I'm getting no other errors. I even tried replacing all occurrances of basic_ostream with just ostringstream and there was no difference (I'll be having a w_char version later, but I wanted the simple case to work first). I made the object on the line above and then passed a fully-constructed object on the line, and the error was exactly the same. I've changed the signature of the second argument to and from const with no change as well.
Any ideas on what I'm doing wrong here?
Edit: since EVERY response seems to want to put it there, I can NOT use std::ostream because I want this to work ONLY for std::ostringstream (and std::basic_ostringstream) and not for any type of output stream. Besides, the function wouldn't compile with ostream anyways, since I'm using the os.str() method, which isn't in ostream, only the sub-classes.
The real problem with your code is that you've overloaded std::ostringstream rather than std::ostream. So your code would work if you write this:
debug_stream_info info(/** blah blah**/);
std::ostringstream oss;
oss << info ; //OK
However this will not work:
oss << 1 << info; //ERROR
This is compilation error because the expression oss<<1 returns an object of type std::ostream& which doesn't have overload which takes debug_stream_info as second argument. That means if you use cast as:
static_cast<std::ostringstream&>(oss << 1) << info; //OK
then that should work again.
So the solution is to overload std::ostream, instead of std::basic_ostringstream.
Also, the second parameter should be const & . This is also a problem with your code.
So write this:
std::ostream& operator<<(std::ostream&, debug_stream_info const &);
//^^^^^^^ note this
The second parameter should be const & so that you could write temporary objects to the stream.
debug_stream_info(__LINE__, __FILE__, LOG_LEVEL); is creating unnamed object which is not returning anything hence error
#include <iostream>
#include <cstdio>
#include <sstream>
using namespace std;
class debug_stream_info
{
public:
debug_stream_info(int errorLine, char *errorFile, int level)
:m_errorLine(errorLine), m_errorFile(errorFile), m_logLevel(level)
{
}
friend std::basic_ostringstream<char>& operator<<(std::basic_ostringstream<char>& os, debug_stream_info& debug_info);
std::ostringstream& fun(std::ostringstream& os)
{
os<<"Ashish"<<endl;
return os;
}
private:
int m_errorLine;
std::string m_errorFile;
int m_logLevel;
};
std::basic_ostringstream<char>& operator<<(std::basic_ostringstream<char>& os, debug_stream_info& debug_info)
{
// Write the stream's contents to cpu_debug
// Deleted custom logging function. No errors here though
// Clear the stream for re-use.
// os.str("");
// os.seekp(0);
return os;
}
int main(int argc, char** argv)
{
std::ostringstream myout, test;
myout << "hey there" << " and some more " << "Numbers!!: " << 435 << 54.2 << " that's good for numbers"
<< debug_stream_info(1, "/home/ashish/test", 1).fun(test);
return 0;
}
Nawaz has explained very clearly why you're getting the error. The
usual solution in this case is to define your own stream type, unrelated
to std::istream. Something along the lines of:
class DebugStream
{
std::ostringstring* collector;
public:
template <typename T>
DebugStream& operator<<( T const& value )
{
if ( collector != NULL ) {
*collector << value;
}
return *this;
}
};
There are infinite variations on this; in your case, you could add a
non-template member function for your type; more likely, you'd add a
constructor which took the same arguments:
DebugStream( int lineNumber, std::string const& filename, int logLevel )
: collector( isActive( logLevel ) ? new std::ostringstream : NULL )
{
// Initial insertion of lineNumber, filename, timestamp...
}
You can also add a destructor which atomically flushes the collected
data to a file (or sends an email, or writes it to the system log, or
whatever). (Be very careful about this. You don't want an exception to
escape from the destructor, even if the logging fails.)
Finally, you might want to use a custom streambuf, rather than
stringstream. Say one that keeps the allocated buffer from one
instance to the next. And if you do this, rather than newing the
stream each time, you might pick up an instance from a table, indexed by
the log level (and initialized from a configuration file).
I would like to be able to do:
foo(stringstream()<<"number = " << 500);
EDIT: single line solution is crucial since this is for logging purposes. These will be all around the code.
inside foo will print the string to screen or something of the sort.
now since stringstream's operator<< returns ostream&, foo's signature must be:
foo(ostream& o);
but how can I convert ostream& to string? (or char*).
Different approaches to achieving this use case are welcome as well.
The obvious solution is to use dynamic_cast in foo. But the given
code still won't work. (Your example will compile, but it won't do what
you think it should.) The expression std::ostringstream() is a
temporary, you can't initialize a non-const reference with a temporary,
and the first argument of std::operator<<( std::ostream&, char const*)
is a non-const reference. (You can call a member function on a
temporary. Like std::ostream::operator<<( void const* ). So the code
will compile, but it won't do what you expect.
You can work around this problem, using something like:
foo( std::ostringstream().flush() << "number = " << 500 );
std::ostream::flush() returns a non-const reference, so there are no
further problems. And on a freshly created stream, it is a no-op.
Still, I think you'll agree that it isn't the most elegant or intuitive
solution.
What I usually do in such cases is create a wrapper class, which
contains it's own std::ostringstream, and provides a templated
member operator<< which forwards to the contained
std::ostringstream. Your function foo would take a const
reference to this—or what I offen do is have the destructor call
foo directly, so that the client code doesn't even have to worry about
it; it does something like:
log() << "number = " << 500;
The function log() returns an instance of the wrapper class (but see
below), and the (final) destructor of this class calls your function
foo.
There is one slight problem with this. The return value may be copied,
and destructed immediately after the copy. Which will wreck havoc with
what I just explained; in fact, since std::ostringstream isn't
copyable, it won't even compile. The solution here is to put all of the
actual logic, including the instance of std::ostringstream and the
destructor logic calling foo in a separate implementation class, have
the public wrapper have a boost::shared_ptr to it, and forward. Or
just reimplement a bit of the shared pointer logic in your class:
class LogWrapper
{
std::ostringstream* collector;
int* useCount;
public:
LogWrapper()
: collector(new std::ostringstream)
, useCount(new int(1))
{
}
~LogWrapper()
{
-- *useCount;
if ( *useCount == 0 ) {
foo( collector->str() );
delete collector;
delete useCount;
}
}
template<typename T>
LogWrapper& operator<<( T const& value )
{
(*collector) << value;
return *this;
}
};
Note that it's easy to extend this to support optional logging; just
provide a constructor for the LogWrapper which sets collector to
NULL, and test for this in the operator<<.
EDITED:
One other thing occurs to me: you'll probably want to check whether the
destructor is being called as a result of an exception, and not call
foo in that case. Logically, I'd hope that the only exception you
might get is std::bad_alloc, but there will always be a user who
writes something like:
log() << a + b;
where the + is a user defined overload which throws.
I would suggest you to use this utility struct:
struct stringbuilder
{
std::stringstream ss;
template<typename T>
stringbuilder & operator << (const T &data)
{
ss << data;
return *this;
}
operator std::string() { return ss.str(); }
};
And use it as:
void f(const std::string & s );
int main()
{
char const *const pc = "hello";
f(stringbuilder() << '{' << pc << '}' );
//this is my most favorite line
std::string s = stringbuilder() << 25 << " is greater than " << 5 ;
}
Demo (with few more example) : http://ideone.com/J995r
More on my blog : Create string on the fly just in one line
You could use a proxy object for this; this is a bit of framework, but if you want to use this notation in a lot of places then it may be worth it:
#include <iostream>
#include <sstream>
static void foo( std::string const &s )
{
std::cout << s << std::endl;
}
struct StreamProxy
{
std::stringstream stream;
operator std::string() { return stream.str(); }
};
template <typename T>
StreamProxy &operator<<( StreamProxy &s, T v )
{
s.stream << v;
return s;
}
static StreamProxy make_stream()
{
return StreamProxy();
}
int main()
{
foo( make_stream() << "number = " << 500 );
}
This program prints
number = 500
The idea is to have a little wrapper class which can be implicitely converted into a std::string. The << operator is simply forwarded to the contained std::stringstream. The make_stream() function is strictly speaking not necessary (you could also say StreamProxy(), but I thought it looks a bit nicer.
A couple of options other than the nice proxy solution just presented by Frerich Raabe:
Define a static string stream variable in the header that defines the logging function and use the comma operator in your invocation of the logging function so that this variable is passed rather than the ostream& returned by the stream insertion operator. You can use a logging macro to hide this ugliness. The problem with this solution is that it is a bit on the ugly side, but this is a commonly used approach to logging.
Don't use C++ I/O. Use a varargs C-style solution instead. Pass a format string as the first argument, with the remaining arguments being targets for that format string. A problem with this solution is that even if your compiler is smart enough to ensure that printf and its cousins are safe, the compiler probably won't know that this new function is a part of the printf family. Nonetheless, this is also a commonly used approach.
If you don't mind using macros functions, you can make the logging function accept const string&, and use the following macro
#define build_string(expr) \
(static_cast<ostringstream*>(&(ostringstream().flush() << expr))->str())
And suppose you foo has signature void foo(const string&), you only need the one-liner
foo(build_string("number = " << 500))
This was inspired by James Kanze's answer about static_cast and stringstream.flush. Without the .flush() the above method fails with unexpected output.
Please note that this method should not leak memory, as temporary values, whether in the pointer form or not, are still allocated on the stack and hence destroyed upon return.
Since you're converting to string anyways, why not
void foo(const std::string& s)
{
std::cout << "foo: " << s << std::endl;
}
...
std::stringstream ss;
ss << "number = " << 500;
foo(ss.str());
This is not possible. As the name ostream implies, it is used for output, for writing to it. You could change the parameter to stringstream&. This class has the method str() which returns a std::string for your use.
EDIT I did not read the issue with operator << returning ostream&. So I guess you cannot simply write your statements within the functions argument list but have to write it before.
You can create a small wrapper around std::ostringstream that will convert back to std::string on use, and have the function take a std::string const &. The first approach to this solution can be found in this answer to a different question.
On top of that, you can add support for manipulators (std::hex) if needed.