C++ Get name of type in template

Question

I'm writing some template classes for parseing some text data files, and as such it is likly the great majority of parse errors will be due to errors in the data file, which are for the most part not written by programmers, and so need a nice message about why the app failed to load e.g. something like:

Error parsing example.txt. Value ("notaninteger")of [MySectiom]Key is not a valid int

I can work out the file, section and key names from the arguments passed to the template function and member vars in the class, however I'm not sure how to get the name of the type the template function is trying to convert to.

My current code looks like, with specialisations for just plain strings and such:

template<typename T> T GetValue(const std::wstring &section, const std::wstring &key)
{
    std::map<std::wstring, std::wstring>::iterator it = map[section].find(key);
    if(it == map[section].end())
        throw ItemDoesNotExist(file, section, key)
    else
    {
        try{return boost::lexical_cast<T>(it->second);}
        //needs to get the name from T somehow
        catch(...)throw ParseError(file, section, key, it->second, TypeName(T));
    }
}

Id rather not have to make specific overloads for every type that the data files might use, since there are loads of them...

Also I need a solution that does not incur any runtime overhead unless an exception occurs, i.e. a completely compile time solution is what I want since this code is called tons of times and load times are already getting somewhat long.

EDIT: Ok this is the solution I came up with:

I have a types.h containg the following

#pragma once
template<typename T> const wchar_t *GetTypeName();

#define DEFINE_TYPE_NAME(type, name) \
    template<>const wchar_t *GetTypeName<type>(){return name;}

Then I can use the DEFINE_TYPE_NAME macro to in cpp files for each type I need to deal with (eg in the cpp file that defined the type to start with).

The linker is then able to find the appropirate template specialisation as long as it was defined somewhere, or throw a linker error otherwise so that I can add the type.

Answer 1

Jesse Beder's solution is likely the best, but if you don't like the names typeid gives you (I think gcc gives you mangled names for instance), you can do something like:

template<typename T>
struct TypeParseTraits;

#define REGISTER_PARSE_TYPE(X) template <> struct TypeParseTraits<X> \
    { static const char* name; } ; const char* TypeParseTraits<X>::name = #X


REGISTER_PARSE_TYPE(int);
REGISTER_PARSE_TYPE(double);
REGISTER_PARSE_TYPE(FooClass);
// etc...

And then use it like

throw ParseError(TypeParseTraits<T>::name);

EDIT:

You could also combine the two, change name to be a function that by default calls typeid(T).name() and then only specialize for those cases where that's not acceptable.

Answer 2

The solution is

typeid(T).name()

which returns std::type_info.

Answer 3

typeid(T).name() is implementation defined and doesn't guarantee human readable string.

Reading cppreference.com :

Returns an implementation defined null-terminated character string containing the name of the type. No guarantees are given, in particular, the returned string can be identical for several types and change between invocations of the same program.

...

With compilers such as gcc and clang, the returned string can be piped through c++filt -t to be converted to human-readable form.

But in some cases gcc doesn't return right string. For example on my machine I have gcc whith -std=c++11 and inside template function typeid(T).name() returns "j" for "unsigned int". It's so called mangled name. To get real type name, use abi::__cxa_demangle() function (gcc only):

#include <string>
#include <cstdlib>
#include <cxxabi.h>

template<typename T>
std::string type_name()
{
    int status;
    std::string tname = typeid(T).name();
    char *demangled_name = abi::__cxa_demangle(tname.c_str(), NULL, NULL, &status);
    if(status == 0) {
        tname = demangled_name;
        std::free(demangled_name);
    }   
    return tname;
}

Answer 4

As mentioned by Bunkar typeid(T).name is implementation defined.

To avoid this issue you can use Boost.TypeIndex library.

For example:

boost::typeindex::type_id<T>().pretty_name() // human readable

Answer 5

The answer of Logan Capaldo is correct but can be marginally simplified because it is unnecessary to specialize the class every time. One can write:

// in header
template<typename T>
struct TypeParseTraits
{ static const char* name; };

// in c-file
#define REGISTER_PARSE_TYPE(X) \
    template <> const char* TypeParseTraits<X>::name = #X

REGISTER_PARSE_TYPE(int);
REGISTER_PARSE_TYPE(double);
REGISTER_PARSE_TYPE(FooClass);
// etc...

This also allows you to put the REGISTER_PARSE_TYPE instructions in a C++ file...

Answer 6

As a rephrasing of Andrey's answer:

The Boost TypeIndex library can be used to print names of types.

Inside a template, this might read as follows

#include <boost/type_index.hpp>
#include <iostream>

template<typename T>
void printNameOfType() {
    std::cout << "Type of T: " 
              << boost::typeindex::type_id<T>().pretty_name() 
              << std::endl;
}

Answer 7

I just leave it there. If someone will still need it, then you can use this:

template <class T>
bool isString(T* t) { return false;  } // normal case returns false

template <>
bool isString(char* t) { return true; }  // but for char* or String.c_str() returns true
.
.
.

This will only CHECK type not GET it and only for 1 type or 2.

Answer 8

If you'd like a pretty_name, Logan Capaldo's solution can't deal with complex data structure: REGISTER_PARSE_TYPE(map<int,int>) and typeid(map<int,int>).name() gives me a result of St3mapIiiSt4lessIiESaISt4pairIKiiEEE

There is another interesting answer using unordered_map or map comes from https://en.cppreference.com/w/cpp/types/type_index.

#include <iostream>
#include <unordered_map>
#include <map>
#include <typeindex>
using namespace std;
unordered_map<type_index,string> types_map_;

int main(){
    types_map_[typeid(int)]="int";
    types_map_[typeid(float)]="float";
    types_map_[typeid(map<int,int>)]="map<int,int>";

    map<int,int> mp;
    cout<<types_map_[typeid(map<int,int>)]<<endl;
    cout<<types_map_[typeid(mp)]<<endl;
    return 0;
}