error compiling lua bridge - visual-c++

i was trying the compile lua bridge from the repository
https://github.com/vinniefalco/LuaBridge/releases
However was getting the error
C2953 'luabridge::FuncTraits': class template has already been defined LuaBridgeDemo luabridgedemo-1.0\luabridgedemo-1.0\luabridge\luabridge.h 1436
Upon closer examination there are two similar structures declared on the header file
template <typename R, typename D>
struct FuncTraits <R (*) () THROWSPEC, D>
{
static bool const isMemberFunction = false;
typedef D DeclType;
typedef R ReturnType;
typedef None Params;
static R call (DeclType fp, TypeListValues <Params> const&)
{
return fp ();
}
};
template <class T, typename R, typename D>
struct FuncTraits <R (T::*) () const THROWSPEC, D>
{
static bool const isMemberFunction = true;
static bool const isConstMemberFunction = true;
typedef D DeclType;
typedef T ClassType;
typedef R ReturnType;
typedef None Params;
static R call (T const* const obj, DeclType fp, TypeListValues <Params> const&)
{
(void)tvl;
return (obj->*fp)();
}
};
I am using Visual C++ 2015. Is there any setting or code change that I need to do to resolve this error.
Thanks in advance

I just removed the duplicate method signatures, it compiled after that.

Related

C++ std::experimental::is_detected not working in MSVC

I have some code which allow a value to be converted into a string, and this works perfectly in g++ and CLion, however when I try to run the same program with MSVC in Visual Studio the program gives many errors, some of which are syntax errors which is quite odd.
This is the code I am using:
// 1- detecting if std::to_string is valid on T
template<typename T>
using std_to_string_expression = decltype(std::to_string(std::declval<T>()));
template<typename T>
constexpr bool has_std_to_string = is_detected<std_to_string_expression, T>;
// 2- detecting if to_string is valid on T
template<typename T>
using to_string_expression = decltype(to_string(std::declval<T>()));
template<typename T>
constexpr bool has_to_string = is_detected<to_string_expression, T>;
// 3- detecting if T can be sent to an ostringstream
template<typename T>
using ostringstream_expression = decltype(std::declval<std::ostringstream&>() << std::declval<T>());
template<typename T>
constexpr bool has_ostringstream = is_detected<ostringstream_expression, T>;
// -----------------------------------------------------------------------------
// 1- std::to_string is valid on T
template<typename T, typename std::enable_if<has_std_to_string<T>, int>::type = 0>
std::string toString(T const& t) {
return std::to_string(t);
}
// 2- std::to_string is not valid on T, but to_string is
template<typename T, typename std::enable_if<!has_std_to_string<T> && has_to_string<T>, int>::type = 0>
std::string toString(T const& t) {
return to_string(t);
}
// 3- neither std::string nor to_string work on T, let's stream it then
template<typename T, typename std::enable_if<!has_std_to_string<T> && !has_to_string<T> && has_ostringstream<T>, int>::type = 0>
std::string toString(T const& t) {
std::ostringstream oss;
oss << t;
return oss.str();
}
I wonder if I am doing something very obviously wrong, or if there is something a bit more complicated leading to the issue. What do I need to change in order to make this program work in Visual Studio and compile correctly?
std::experimental::is_detected is not supported in Visual Studio 2019 or earlier. You can pretty easily write your own cross platform implementation of it though, as demonstrated here: https://topanswers.xyz/cplusplus?q=1295

c++ parse from string using template

In C++, I want to get value from string.
I know there are functions like stoi, stol, stof...
But can I write a function that includes all of them? like using a template to infer the data type?
template<typename T>
T fromString(const std::string& s){
// do something here
// call stoi, stof according to typename T
}
string si = "1234";
int integer = fromString<int>(si);
string sf = "1234.1234";
float float_point = fromString<float>(sf);
Cannot comment yet, hence this ...
Not clear if you're looking for a single function, or templated (overloaded) functions? This would work: Have no default implementation with return type T, and specializations / usage so:
template<typename T> foo(const std::string& str) {}
template<> int foo(const std::string& str) {return atoi(str.c_str());}
template<> double foo(const std::string& str) {return atof(str.c_str());}
...
const std::string str1("1234"); const std::string str2("12.34");
const auto iVal = foo<int>(str1);
const auto dVal = foo<double>(str2);
Note the absence of specialization in the function name with the template function specialization, of course it appears as a return type.

Auto-interpreting a c-style string as a std::string via Boost's Property tree's .get function

I use boosts property tree, included via
#include "boost\property_tree\ptree.hpp"
And... I'd like to create a simple function which substitutes a value in case none is found via a fairly straight-forward template function:
template <typename Type>
Type getValueOrDefault( std::string const& str, Type defaultValue )
{
Type returnValue = defaultValue;
try {
returnValue = mSettings.get<Type>( str );
}
catch ( boost::property_tree::ptree_error &e )
{
// Log error!
}
return returnValue;
}
This works well in principle, but runs into a bit problems if I rely on C-style string. For example, calling the function as follows:
getValueOrDefault( "pathToImportantStuffParameter", "c:/defaultdir/" )
will result in the following error:
boost\property_tree\stream_translator.hpp(36): error C2678: binary '>>' : no operator found which takes a left-hand operand of type 'std::basic_istream<char,std::char_traits<char>>' (or there is no acceptable conversion)
The error stems from passing char const * as a template parameter which makes a fair bit of sense. Two obvious solutions to this issue would be to force the default value to be a std::string object, like so:
getValueOrDefault<std::string>( "pathToImportantStuffParameter", "c:/defaultdir/" )
getValueOrDefault( "pathToImportantStuffParameter", std::string("c:/defaultdir/") )
But I'm wondering if someone might know of some template magic I could sprinkle to automatically interpret c-style strings as std::strings?
You can provide a char array overload which converts the char array to a std::string and then calls the default implementation:
#include <iostream>
#include <string>
template <typename T>
T getValueOrDefault(const std::string& str, T&& defaultValue)
{
std::cout << "inside default implementation" << std::endl;
/* ... */
return defaultValue;
}
template <std::size_t N>
std::string getValueOrDefault(const std::string& str, const char (&defaultValue)[N])
{
std::cout << "inside char[] overload" << std::endl;
return getValueOrDefault(str, std::string(defaultValue));
}
int main()
{
auto x = getValueOrDefault("foo", "bar");
return 0;
}
live example
An alternative solution is to use custom type traits:
#include <string>
#include <type_traits>
template <typename T>
struct return_type
{
using type = T;
};
template <>
struct return_type<const char*>
{
using type = std::string;
};
template <typename T>
using return_type_t = typename return_type<typename std::decay<T>::type>::type;
template <typename T>
return_type_t<T> getValueOrDefault(const std::string& str, T&& defaultValue)
{
return_type_t<T> value(defaultValue);
/* ... */
return value;
}
int main()
{
auto x = getValueOrDefault("foo", "bar");
static_assert(std::is_same<decltype(x), std::string>::value, "");
return 0;
}
live example
The only way I found is to specialize getValueOrDefault for const char*, which calls getValueOrDefault with std::string explicitly:
//Note that the return value is unspecified, it returns a 'const char*' to a temporary,
//which will be destroyed when the function returns
template <>
const char* getValueOrDefault(std::string const& str, const char* defaultValue)
{
return getValueOrDefault<std::string>(str, defaultValue).c_str();
}
If you want to that function to return a std::string instead of an invalid const char*, you have to change the template signature a bit:
//Default return type is the same as paramter
template <typename Type, typename Return = Type>
Return getValueOrDefault(std::string const& str, Type defaultValue)
{
//...
}
//Trick the compiler to select this overload for 'const char*'
template <typename Return = std::string>
Return getValueOrDefault(std::string const& str, const char* defaultValue)
{
return getValueOrDefault<std::string, std::string>(str, defaultValue);
}
or you could just plain overload the function (thanks #m.s.)
//Overload for 'const char*'
std::string getValueOrDefault(std::string const& str, const char* defaultValue)
{
return getValueOrDefault<std::string>(str, defaultValue);
}
There is also a third way (if you can use C++14), using the string literal ""s:
//"c:/defaultdir/"s is a std::string (note the s after it => string literal)
getValueOrDefault("pathToImportantStuffParameter", "c:/defaultdir/"s);

Fitting string literals for different string classes

The problem
I am implementing a class where I want to let the user choose the string type (std::string, std::wstring, std::u16string, ...) via a template parameter. I currently fail to make the string literals fit the chosen string type: Once I decide for a literal prefix ("hello" vs. L"hello" vs. u"hello" vs. U"hello"), I get compilation errors for all incompatible string classes.
Toy example
As an example, consider the following code (compile with --std=c++11):
#include <string>
template<typename StringType>
void hello_string()
{
StringType result("hello");
}
int main()
{
// works
hello_string<std::string>();
hello_string<std::basic_string<char>>();
// the code below does not compile
hello_string<std::wstring>();
hello_string<std::basic_string<unsigned char>>();
hello_string<std::u16string>();
}
Function hello_string() shows the essence of what I want to do: have a string type as template parameter, and assign string literals to variables of this type.
Possible workaround
One way to overcome my problem would be to implement several specializations of the hello_string() function. The problem is that this would lead to several copies of each string literal - one for each string literal prefix. I think this is rather ugly, and there must be better way.
Another way could be to chose "normal" string literals as default values and have functions do a conversion to the different string types. While this would avoid code duplication, it would introduce unnecessary conversions of something that is actually constant.
You can make yourself a macro. First define a struct that wraps the char-choosing:
namespace details {
template<typename T>
struct templ_text;
template<>
struct templ_text <char>
{
typedef char char_type;
static const char_type * choose(const char * narrow, const wchar_t * wide, const char16_t* u16, const char32_t* u32) { return narrow; }
static char_type choose(char narrow, wchar_t wide, char16_t u16, char32_t u32) { return narrow; }
};
template<>
struct templ_text < wchar_t >
{
typedef wchar_t char_type;
static const char_type* choose(const char * narrow, const wchar_t * wide, const char16_t* u16, const char32_t* u32) { return wide; }
static char_type choose(char narrow, wchar_t wide, char16_t u16, char32_t u32) { return wide; }
};
template<>
struct templ_text < char16_t >
{
typedef char16_t char_type;
static const char_type* choose(const char * narrow, const wchar_t * wide, const char16_t* u16, const char32_t* u32) { return u16; }
static char_type choose(char narrow, wchar_t wide, char16_t u16, char32_t u32) { return u16; }
};
template<>
struct templ_text < char32_t >
{
typedef char32_t char_type;
static const char_type* choose(const char * narrow, const wchar_t * wide, const char16_t* u16, const char32_t* u32) { return u32; }
static char_type choose(char narrow, wchar_t wide, char16_t u16, char32_t u32) { return u32; }
};
}
Wrap it into a nice macro:
#define TEMPL_TEXT(Ch, txt) details::templ_text<Ch>::choose(txt, L##txt, u##txt, U##txt)
Then your function would be:
template<typename StringType>
void hello_string()
{
StringType result(TEMPL_TEXT(typename StringType::value_type, "Hello"));
}
I think that unused copies of the string will be optimized away.

How is OpenCL floatn different from a custom struct

I'm having problems passing a structure of vectors to an OpenCL kernel.
Long story short, I have found that I CAN pass a C structure:
typedef struct {
cl_float4 vec;
} my_type;
. . . to a kernel, if I declare it like:
typedef struct { float s[4]; } my_float4;
typedef struct { my_float4 vec; } my_type; //use custom float4
__kernel void function(const my_type test) {}
. . . but NOT, if I declare it like:
typedef struct { float4 vec; } my_type; //use built-in float4
__kernel void function(const my_type test) {}
So my question is, what makes a float4 different and special? In the second example, I get CL_INVALID_ARG_SIZE on clSetKernelArg. In the host code, I'm using cl_float4--isn't that supposed to be paired with float4 in the kernel?
EDIT: by request, the clSetKernelArg code (error handling cropped):
template <typename type_data> void set_argument(int arg_index, const type_data* data) {
cl_int err = clSetKernelArg(kernel,arg_index,sizeof(type_data),data);
if (err!=CL_SUCCESS) throw "ERROR"; /*simplified*/
}
I call the function with a struct declared exactly as above:
my_type test;
my_kernel->set_argument(0,&test);
This sounds like a bug with your implementation - you should be able to pass that user-defined type to a kernel without issue. What platform / SDK are you using? I was able to successfully declare this type in host code:
typedef struct _mytype {
cl_float4 val;
} mytype;
... and then later use it:
mytype t;
status = clSetKernelArg(kernel, argc++, sizeof(mytype), &t);
check_status("clSetKernelArg:mytype t", status);
Where check_status does what you'd expect: checks for a non-CL_SUCCESS return value.
The kernel looks like this:
typedef struct _mytype {
float4 value;
} mytype;
kernel void do_it(/* ... */, const mytype test) { /* ... */ }
This compiles and executes, giving correct results on the CPU device.

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