Use boost::optional together with boost::adaptors::indirected - c++

I am trying to compile the following code:
#include <iostream>
#include <iterator>
#include <vector>
#include <boost/assign/std/vector.hpp>
#include <boost/optional.hpp>
#include <boost/range/adaptor/indirected.hpp>
#include <boost/range/algorithm/copy.hpp>
int main( int argc, char ** argv )
{
using namespace boost::assign;
using boost::adaptors::indirected;
std::vector<boost::optional<unsigned> > values;
values += 1u,2u,3u;
boost::copy( values | indirected, std::ostream_iterator<unsigned>( std::cout, " " ) );
std::cout << std::endl;
}
However, I got some errors, e.g. that there is no type named element_type in boost::optional<unsigned>. The reference page page, however, says that the single precondition is the existence of the operator*() unary function. Is there a way to make it work?

This is definitely a bug in Boost, but whether that bug is in Boost.Optional or Boost.Iterator is up for debate (I would say the latter, personally).
However, the fix is trivial -- before including any Boost headers, do this:
#include <boost/optional/optional_fwd.hpp>
#include <boost/pointee.hpp>
namespace boost
{
template<typename P>
struct pointee<optional<P> >
{
typedef typename optional<P>::value_type type;
};
}
Then include other Boost headers as necessary.
Please submit a ticket on the Boost Trac, or at the least post a bug report on the Boost Users mailing list.

Look at the private optional.hpp defined in boost iostreams library here. You will see that it defines a typedef T element_type;
However the actual optional.hpp that you are using defined here does not define it. So that is why the compiler is complaining. I don't know why it was overlooked.
Try using the private optional.hpp from iostreams library to solve this issue. I hope this helps.

Related

Application crash when using library from the program compiled with make

I'm making scientific calculator in command line in c++ for my usage and also for practice. I have a problem with compiling it using cmake with mingw on windows. These are my source files:
main.ccp
#include <iostream>
#include <string>
#include "ExpressionCalculations/ExpressionParser.h"
int main()
{
std::string humanReadableExpression;
std::cout<<"Enter expression\n";
std::getline(std::cin, humanReadableExpression);
std::cout<<humanReadableExpression;
ExpressionCalculations::ExpressionParser parser;
auto&& expression = parser.GenerateRpnExpression(humanReadableExpression);
return 0;
}
ExpressionParser.h
#pragma once
#include <memory>
#include <stack>
#include <string>
#include <unordered_map>
namespace ExpressionCalculations
{
class ExpressionParser
{
public:
std::unique_ptr<std::string> GenerateRpnExpression(std::string &humanReadableExpression);
private:
// other code
};
}
ExpressionParser.cpp
#include <memory>
#include <stack>
#include <string>
#include <unordered_map>
#include <iostream>
#include "ExpressionParser.h"
namespace ExpressionCalculations
{
std::unique_ptr<std::string> ExpressionParser::GenerateRpnExpression(
std::string& humanReadableExpression)
{
std::unique_ptr<std::string> rpnExpression;
*rpnExpression="3456";
return rpnExpression;
}
These are cmake files
main CMakeLists.txt
cmake_minimum_required (VERSION 2.8)
project (ScientificCalculator_exe)
add_subdirectory(ExpressionCalculations)
add_executable(ScientificCalculator main.cpp)
target_link_libraries(ScientificCalculator ExpressionCalculations)
module CMakeList.txt
set(calculators ExpressionParser.h ExpressionParser.cpp)
add_library(ExpressionCalculations ${calculators})
When I run it , I can see Enter expression and pass input. Then I get Segmentation fault. However when I remove declaration of ExpressionParser and auto&& expression the string is shown, a string can be inputted and shown in the command. I checked configuration question multiple directories under cmake, https://cmake.org/cmake-tutorial/ and https://www.codeproject.com/Articles/1181455/A-CMake-tutorial-for-Visual-Cplusplus-developers but it seems that I correctly made cmake files. I have no idea why it doesn't work. I use the latest mingw64 on windows with default make compilation parameters.
From the cppreference page on unique_ptr:
The class satisfies the requirements of MoveConstructible and MoveAssignable, but not the requirements of either CopyConstructible or CopyAssignable.
In your ExpressionParser::GenerateRpnExpression function you are attempting to copy the unique_ptr out of the function when you should be moving it. Try return std::move(rpnExpression)
After debugging program compiled with just g++ I found the problem. It was misunderstanding of unique_ptr' default constructor behaviour. I thought it would initialize std::string but after reading doc and checking it it does not initialize object and generates nullptr. Then I looked into Scott Myers's Modern Effective C++ how to initialize the unique_ptr. Instead of std::unique_ptr<std::string> rpnExpression; I used auto rpnExpression = std::make_unique<std::string>();. It works like charm. I checked compiling through cmake and there were not any problems.

Using Boost::iostreams dual-use filters

I was attempting to follow the example of Finite State Filters in the Boost::iostreams documentation. However when I went to use the filter I got an error stating the ::imbue was not accessible because 'boost::iostreams::detail::finite_state_filter_impl' uses 'protected' to inherit from 'my_fsm'.
Frustrated I copied my code into the tests used to in the boost examples. The tests compile and pass. My conculsion is that I am probably mis-using the dual use filter defined by:
typedef io::finite_state_filter my_fsm_filter;
I feel that just pushing it onto a filtered_stream may not be proper, but I could not find a missing step. I am sure there must be a need to wrap the filter but I can find no example (though I am sure if I dug deep enough into the code used to test the boost code it has to be there somewhere).
here is a bit of example code:
#include <boost/mpl/vector.hpp>
#include <libs/iostreams/example/finite_state_filter.hpp>
namespace io = boost::iostreams;
struct my_fsm : io::finite_state_machine<my_fsm> {
BOOST_IOSTREAMS_FSM(my_fsm) // define skip and push.
typedef my_fsm self;
static const int beginline = 0;
static const int skipline = 1;
static const int dataline = 2;
typedef boost::mpl::vector <
row<beginline, is<'C'>, skipline, &self::skip>,
row<beginline, is_any, dataline, &self::push>,
row<skipline, is<'\n'>, beginline, &self::skip>,
row<skipline, is_any, skipline, &self::skip>,
row<dataline, is<'\n'>, beginline, &self::push>,
row<dataline, is_any, dataline, &self::push>
> transition_table;
};
typedef io::finite_state_filter<my_fsm> my_fsm_filter;
#include <iostream>
#include <string>
#include <boost/iostreams/device/file.hpp>
#include <boost/iostreams/filtering_stream.hpp>
#include <boost/iostreams/stream.hpp>
namespace io = boost::iostreams;
int main() {
io::stream<io::file_sink> out(io::file_sink("outputfile.txt"));
io::filtering_istream in;
my_fsm_filter infsm;
in.push(my_fsm_filter());
in.push(io::file_source("inputdata.txt"));
while (in) {
std::string line;
if(std::getline(in, line)) {
//std::cout << line << std::endl;
out << line << std::endl;
}
}
return 0;
}
I personally feel that there is a bug in the sample header with respect to this imbue call.
However, you can work around it by changing the typedef to
struct my_fsm_filter : io::finite_state_filter<my_fsm> {
using io::finite_state_filter<my_fsm>::imbue;
};
This explicitly exposes the imbue method as public on the derived type. I haven't looked at the sample program that you reported to be working (because you didn't link to it). But it's possible they used a similar hack.
In my tests, a similar edit to finite_state_filte.hpp L278 to add
using base_type::imbue;
to class finite_state_filter has the same effect.

Boost Phoenix compilation error

I was trying using Boost Phoenix. My aim is to have stl algorithms that take a container instead of range of iterators, as described
here.
However, I am getting a mess of errors on a rather simple code:
#include <boost/phoenix/stl/algorithm.hpp>
#include <iostream>
#include <vector>
#include <algorithm>
namespace phx = boost::phoenix;
int main(int argc, char ** argv) {
std::vector<int> my_arr{ 0, 1, 2, 3 };
phx::for_each(my_arr, [](int i){std::cout << i << std::endl; });
}
I tried it on two platforms (Win7, Ubuntu) where I otherwise use multiple parts of Boost.
The error messages are rather long so I put them in files:
MVC++ November 2013, Boost 1.55
G++4.7.2, Boost 1.53
I can't really make much of the errors apart from templates not being compiled correctly, but I guess I am missing something rather simple.
I think you are using the wrong boost library. The phoenix algorithms are lazy functions which are explained here. So phoenix::for_each does not do anything if it is called, but returns a function object which iterates over the range once it is called. If you simply want to use the STL (and other) algorithms on ranges you can use the boost.range library:
#include <boost/range/algorithm/for_each.hpp>
#include <iostream>
#include <vector>
namespace rng = boost::range;
int main(int argc, char ** argv) {
std::vector<int> my_arr{ 0, 1, 2, 3 };
rng::for_each(my_arr, [](int i){std::cout << i << std::endl; });
}
You simply need to include the phoenix core before including anything else.
#include <boost/phoenix/core.hpp>
#include <boost/phoenix/stl/algorithm.hpp>
... rest of your program

Boost-spirit-karma and boost-variant "concepts" related to auto generators

I need to deserialize a std::vector<boost::variant<..>> with decoration supplied by other objects.
One of the things the "decoration" enables is a empty entry in the vector. I have hit a brick wall in my real implementation. However, I have managed to shrink wrap it. Code that compiles :
#include <string>
#include <boost/spirit/include/karma.hpp>
#include <boost/variant.hpp>
#include <boost/cstdint.hpp>
namespace karma = boost::spirit::karma;
typedef boost::variant<boost::int32_t, boost::int64_t> custom_variant;
int main()
{
using karma::generate;
custom_variant v;
std::string temp;
std::back_insert_iterator<std::string> x(temp);
std::cout << v;
karma::generate(x, karma::auto_, v);
}
The offending changes, which attempt to implement a "undefined" type, along with the required concept.
#include <string>
#include <boost/spirit/include/karma.hpp>
#include <boost/variant.hpp>
#include <boost/cstdint.hpp>
namespace karma = boost::spirit::karma;
struct undefined{};
std::ostream & operator<<(std::ostream & out, undefined const & undefined)
{
return out;
}
typedef boost::variant<undefined,boost::int32_t, boost::int64_t> custom_variant;
int main()
{
using karma::generate;
custom_variant v;
std::string temp;
std::back_insert_iterator<std::string> x(temp);
std::cout << v;
karma::generate(x, karma::auto_, v);
}
If I comment out the karma::generate step, std::cout is a valid expression (Boost::variant OutputStreamable). Spirit requires that generators be given types which are OutputStreamable(spirit::karma OutputStreamable) and the variant above should be OutputStreamable since I have made the undefined typeOutputStreamable as a no-op.
What gives ? :(
I'm really beginning to question weather the C++ template mechanism is worth it when using libraries with > 2 levels of template indirection. Perhaps I should go back to straight-c.
Edit 1:
Ok, Clang gave me a sensible first error...
error: no type named 'properties' in 'boost::spirit::karma::no_auto_mapping_exists'
Now I got to figure out how to map undefined as a no-op to get a clean conversion. This spirit documentation entry (and this in specific) describes what I need to look into. Is there a generic undefined type provided by spirit or one defined in boost, that spirit already maps as a no-op ?
Edit 2:
std::vector<boost::optional<boost::variant<..>>> is beginning to look quite appealing since spirit provides type-deduction for them.
I'd suggest to use spirit::unused_type for that purpose as it already is 'known' to Spirit and it has an operator<<() predefined (but any other type will do) - not that you really need that operator for Karma in the first place.
In addition, you have to provide a specialization for create_generator (as you suspected):
namespace boost { namespace spirit { namespace traits
{
template <>
struct create_generator<spirit::unused_type>
{
typedef spirit::karma::eps_type type;
static type call()
{
return spirit::karma::eps;
}
};
}}}
which will map unused_type to karma::eps. This seems to be exactly what you need as eps eats the attribute without generating anything, while succeeding always. If you go this route you will not need to utilize optional<>.

Does std::map<key, data> in C++ support native data types like Structures?

How do I map a key to a native data type like structure?
I wrote this snipped but I couldn't compile it. Do you have any ideas on how to fix it?
#include <map>
#include <iostream>
typedef struct _list
{
int a,b;
}list;
map<int,list> test_map;
int main(void)
{
cout <<"Testing"<< endl;
}
map resides in the std:: namespace. Two possible ways to fix this:
using namespace std;
// ...
map<int, list> test_map;
or
std::map<int, list> test_map;
I prefer the second method, but it's a purely personal choice.
On a related note, there is no real limitation on what you can put in a map, aside from the fact that they must be copyable/assignable, and that the key type must have a < operator (or you can also provide a comparer functor).
EDIT: Seems like <list> is included somewhere, either in <iostream> (unlikely) or <map> (strange but not impossible). A using namespace std will cause std::list to clash with your own struct. The solution: rename your struct, or remove the using namespace and put std:: where it's needed.
Added std where required.
Renamed list to mylist to avoid clash with std::list. Avoid typenames and variable names that clash with common usage.
Now compiles OK in VS2008.
#include <map>
#include <iostream>
typedef struct _list
{
int a,b;
} mylist;
std::map<int,mylist> test_map;
int main(void)
{
std::cout <<"Testing"<< std::endl;
return 0;
}
There's no issue with using your struct in the STL containers provided it's copyable cleanly (copy constructor), assignable (implements operator=) and comparable (implements operator<).
A number of problems here:
You're missing either a using::std or std::map, so the compiler doesn't know what map<int,list> means.
Assuming you have a using namespace std declaration, your typedef list might collide with the STL collection of the same name. Change the name.
Your typedef struct _tag {...} tag; construct is an archaic holdover from the 80's. It is not necesarry, and frankly rather silly. It gets you nothing.
Here's your code fixed:
#include <map>
#include <iostream>
struct MyList
{
int a,b;
};
std::map<int,MyList> test_map;
int main(void)
{
std::cout <<"Testing"<< std::endl;
return 0;
}
map<int, _list> test_map; or don't use list(much better) as a name of structure. (You probably also have
#include <list>
...
using namespace std;
somewhere in your code.
I would try to avoid using codepad at all.
I have done a couple of tests with your code and it seems that
it is adding an implicit (and unwanted) using namespace std --it does not require you to qualify map, cout or endl.
it is (probably) including more standard headers than you might want, including #include <list>.
That means that when the compiler looks at the code it is seeing two list, your version and the one in std. Because of the using directive, both are in scope in the line where you create the map and the compiler is not able to determine which to use.
Two simple things that you can do: change the name of your type for the simple test to something other than list (ouch! the tool forcing your naming choices!) or fully qualify the use:
#include <map>
struct list {
int a,b;
};
std::map< int, ::list > the_map;
// ...
Note that codepad is adding the include by itself and the using directive, so it will also compile:
struct list {
int a,b;
};
map<int,::list> the_map;
But that piece of code is wrong
You seem to be comming from C. Try this:
#include <map>
#include <iostream>
struct list
{
int a,b;
};
std::map<int,list> test_map;
int main(void)
{
std::cout <<"Testing"<< std::endl;
return 0;
}