Suppose we have enum with implemented operator<< converting enum value to string. Is it possible to call this operator during string construction or something similar? My current approach uses std::stringstream calling << on enum and extracting string from std::stringstream. Is there any other way? Using std::cout << is not an option.
Example code:
enum class Status {
OK,
ERROR
}
::std::ostream& operator<<(::std::ostream& os, Status status)
{
switch (status) {
case Status::OK:
return os << "OK";
case Status::ERROR:
return os << "ERROR";
}
Usage:
Status s = Status::OK;
std::stringstream stream;
stream << s;
std::string statusString = s.str().c_str();
Unfortunately, here is no any elegant solution as you want.
If we only could, we might use a user defined conversion but it needs to be a non-static member function which is impossible with enums.
But you may always use argument-dependent lookup like you do with your operator<<.
Or if you want string, make a map and put desired string-equivalent there.
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
A problem occurred when trying to print out a class method which has en enum class type.
I have a method, which returns an enum which is casted to bool:
bool Window::IsVSync() const
{
return static_cast<bool>(m_WindowData.VSync);
}
m_WindowData.VSync is VSync type.
And that's the enum class:
enum class VSync : bool
{
Disable = false,
Enable = true
}
I know that with enum class I don't get an implicit conversion, and that's the reason I cast to bool. And like that everything is working just fine. But with the fact that my enum class is extended from bool in mind, can I do something like this:
VSync Window::IsVSync() const
{
return m_WindowData.VSync;
}
At first, I thought I could, since we extend the enum class from bool, but I was wrong, and It gives me this error when I try to print to the console:
binary '<<': no operator found which takes a right-hand operand of
type 'VSync' (or there is no acceptable conversion)
I print the method to the console with:
std::cout << myWindow.IsVSync() << "\n";
Which is of course, supposed to return either 1 or 0;
When I cast it to bool in the output like that:
std::cout << static_cast<bool>(myWindow.IsVSync()) << "\n";
it works just fine, but I do not want to make the user have to cast it to bool for it to work. I need a solution which is in the method itself, so the calling stays like that
std::cout << myWindow.IsVSync() << "\n";
You can define a stream operator overload outside of the class for the enum type like so:
static std::ostream& operator<<(std::ostream& os, const VSync& val)
{
switch (val) {
case VSync::Disable:
os << "VSync disabled";
break;
case VSync::Enable:
os << "VSync enabled";
break;
}
return os;
}
In this way you don't need to cast anything and can print something more detailed than a 1 or 0 (or having to use std::boolalpha).
I hope that can help.
It is not extended from bool, is uses the same amount of space as a bool.
With enum, colon does not denote inheritance.
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).
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.
I want to output my own object to a STL stream but with customized formatting. I came up with something like this but since I never used locale and imbue before I have no idea if this makes sense and how to implement MyFacet and operator<<.
So my questions are: does this make sense and how to implement MyFacet and operator<< ?
The following is a simplified example which shows you what I want to do.
struct MyObject
{
int i;
std::string s;
};
std::ostream &operator<<(std::ostream &os, const MyObject &obj)
{
if (????)
{
os << obj.i;
}
else
{
os << obj.s;
}
}
MyObject o;
o.i = 1;
o.s = "hello";
std::cout.imbue(locale("", new MyFacet(MyFacet::UseInt)));
std::cout << o << std::endl; // prints "1"
std::cout.imbue(locale("", new MyFacet(MyFacet::UseString)));
std::cout << o << std::endl; // prints "hello"
Implementing your own operator << for tracing is generally a good idea. However I've never needed to imbue locales. However I tried it and it worked fine. Here's what I did:
class my_facet : public std::locale::facet
{
public:
enum option{
use_string,
use_numeric
};
//Unique id for facet family, no locale can contain two
//facets with same id.
static std::locale::id id;
my_facet(option o=use_numeric):
facet(0),
_option(o)
{//Initialize reference count to zero so that the memory
//management will be handled by locale
};
option get_option() const {return _option;};
protected:
option _option;
};
std::locale::id my_facet::id(123456); //Facet family unique id
std::ostream& operator<<(std::ostream& os, const myobj& o)
{
std::locale const& l = os.getloc();
if( std::has_facet<my_facet>(l) ){
my_facet const& f = std::use_facet<my_facet>(l);
switch(f.get_option()){
case my_facet::use_numeric:
os << "Using numeric! ";
break;
case my_facet::use_string:
os << "Using string! ";
break;
default:
os << "Unhandled case.. ";
break;
}
return os;
}
os << "Default case when no facet has been set";
return os;
}
Then to imbue a locale with the facet:
std::locale mylocale(locale("US"), new my_facet(my_facet::use_numeric));
std::cout.imbue(mylocale);
However a more elegant way would be to implement different facets of the same facet family that can be replaced in the locale.
Well, a locale is generally used to allow different output/input formatting of the same object based on the local (the specified locale in fact) formatting which is present. For a good article on this see: http://www.cantrip.org/locale.html. Now maybe its because your example above is quite simplified, but to me it looks like you are trying to come up with a clever way to switch between printing one part of an object or another. If that is the case it might be simpler do just overload the stream operator for each type and use the if switch externally.
Anyway, I'm not going to pretend that I'm an expert in facets and locales but have a look at that article, its pretty thorough and will give you a better explanation than I will!