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Print/output type info

stackoverflow Is it possible to print a variable's type in standard C++?

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Here is a summary of Howard's long form solution but implemented with a heretical one-line macro: #define DEMANGLE_TYPEID_NAME(x) abi::__cxa_demangle(typeid((x)).name(), NULL, NULL, NULL). If you need cross-platform support: Use #ifdef, #else, #endif to provide one macros for other platforms like MSVC. – Trevor Boyd Smith Jul 5 '16 at 15:01

A

NOTE:

1、这个回答其实展示了各种封装C-string的方式，tag 为 "C string wrapper"

C++11 update to a very old question: Print variable type in C++.

The accepted (and good) answer is to use typeid(a).name(), where a is a variable name.

Now in C++11 we have decltype(x), which can turn an expression into a type. And decltype() comes with its own set of very interesting rules. For example decltype(a) and decltype((a)) will generally be different types (and for good and understandable reasons once those reasons are exposed).

Will our trusty typeid(a).name() help us explore this brave new world?

No.

But the tool that will is not that complicated. And it is that tool which I am using as an answer to this question. I will compare and contrast this new tool to typeid(a).name(). And this new tool is actually built on top of typeid(a).name().

The fundamental issue:

typeid(a).name()

throws away cv-qualifiers, references, and lvalue/rvalue-ness. For example:

const int ci = 0;
std::cout << typeid(ci).name() << '\n';

For me outputs:

i

and I'm guessing on MSVC outputs:

int

I.e. the const is gone. This is not a QOI (Quality Of Implementation) issue. The standard mandates this behavior.

What I'm recommending below is:

template <typename T> std::string type_name();

which would be used like this:

const int ci = 0;
std::cout << type_name<decltype(ci)>() << '\n';

and for me outputs:

int const

<disclaimer> I have not tested this on MSVC. </disclaimer> But I welcome feedback from those who do.

The C++11 Solution

NOTE:

1、理解下面这段代码的前提是理解 __cxa_demangle

Returns:

A pointer to the start of the NUL-terminated demangled name, or NULL if the demangling fails. The caller is responsible for deallocating this memory using free.

即需要由调用者来释放返回的内存。

这让我想起了resource return，从下面作者的封装方式来看，它使用的是"unique_ptr custom deleter-RAII"方式。

2、下面的实现有一个非常不好的是: std::string r = own != nullptr ? own.get() : typeid(TR).name(); 进行了deep copy，其实这一次deep copy是可以不必的。

I am using __cxa_demangle for non-MSVC platforms as recommend by ipapadop in his answer to demangle types. But on MSVC I'm trusting typeid to demangle names (untested). And this core is wrapped around some simple testing that detects, restores and reports cv-qualifiers and references to the input type.

#include <type_traits>
#include <typeinfo>
#ifndef _MSC_VER
#   include <cxxabi.h>
#endif
#include <memory>
#include <string>
#include <cstdlib>

template <class T>
std::string
type_name()
{
    typedef typename std::remove_reference<T>::type TR;
    std::unique_ptr<char, void(*)(void*)> own
           (
#ifndef _MSC_VER
                abi::__cxa_demangle(typeid(TR).name(), nullptr,
                                           nullptr, nullptr),
#else
                nullptr,
#endif
                std::free
           );
    std::string r = own != nullptr ? own.get() : typeid(TR).name();
    if (std::is_const<TR>::value)
        r += " const";
    if (std::is_volatile<TR>::value)
        r += " volatile";
    if (std::is_lvalue_reference<T>::value)
        r += "&";
    else if (std::is_rvalue_reference<T>::value)
        r += "&&";
    return r;
}

The Results

With this solution I can do this:

int& foo_lref();
int&& foo_rref();
int foo_value();

int
main()
{
    int i = 0;
    const int ci = 0;
    std::cout << "decltype(i) is " << type_name<decltype(i)>() << '\n';
    std::cout << "decltype((i)) is " << type_name<decltype((i))>() << '\n';
    std::cout << "decltype(ci) is " << type_name<decltype(ci)>() << '\n';
    std::cout << "decltype((ci)) is " << type_name<decltype((ci))>() << '\n';
    std::cout << "decltype(static_cast<int&>(i)) is " << type_name<decltype(static_cast<int&>(i))>() << '\n';
    std::cout << "decltype(static_cast<int&&>(i)) is " << type_name<decltype(static_cast<int&&>(i))>() << '\n';
    std::cout << "decltype(static_cast<int>(i)) is " << type_name<decltype(static_cast<int>(i))>() << '\n';
    std::cout << "decltype(foo_lref()) is " << type_name<decltype(foo_lref())>() << '\n';
    std::cout << "decltype(foo_rref()) is " << type_name<decltype(foo_rref())>() << '\n';
    std::cout << "decltype(foo_value()) is " << type_name<decltype(foo_value())>() << '\n';
}

and the output is:

decltype(i) is int
decltype((i)) is int&
decltype(ci) is int const
decltype((ci)) is int const&
decltype(static_cast<int&>(i)) is int&
decltype(static_cast<int&&>(i)) is int&&
decltype(static_cast<int>(i)) is int
decltype(foo_lref()) is int&
decltype(foo_rref()) is int&&
decltype(foo_value()) is int

Note (for example) the difference between decltype(i) and decltype((i)). The former is the type of the declaration of i. The latter is the "type" of the expression i. (expressions never have reference type, but as a convention decltype represents lvalue expressions with lvalue references).

Thus this tool is an excellent vehicle just to learn about decltype, in addition to exploring and debugging your own code.

In contrast, if I were to build this just on typeid(a).name(), without adding back lost cv-qualifiers or references, the output would be:

decltype(i) is int
decltype((i)) is int
decltype(ci) is int
decltype((ci)) is int
decltype(static_cast<int&>(i)) is int
decltype(static_cast<int&&>(i)) is int
decltype(static_cast<int>(i)) is int
decltype(foo_lref()) is int
decltype(foo_rref()) is int
decltype(foo_value()) is int

I.e. Every reference and cv-qualifier is stripped off.

C++14 Update

Just when you think you've got a solution to a problem nailed, someone always comes out of nowhere and shows you a much better way. :-)

This answer from Jamboree shows how to get the type name in C++14 at compile time. It is a brilliant solution for a couple reasons:

1、It's at compile time!

2、You get the compiler itself to do the job instead of a library (even a std::lib). This means more accurate results for the latest language features (like lambdas).

Jamboree's answer doesn't quite lay everything out for VS, and I'm tweaking his code a little bit. But since this answer gets a lot of views, take some time to go over there and upvote his answer, without which, this update would never have happened.

#include <cstddef>
#include <stdexcept>
#include <cstring>
#include <ostream>

#ifndef _MSC_VER
#  if __cplusplus < 201103
#    define CONSTEXPR11_TN
#    define CONSTEXPR14_TN
#    define NOEXCEPT_TN
#  elif __cplusplus < 201402
#    define CONSTEXPR11_TN constexpr
#    define CONSTEXPR14_TN
#    define NOEXCEPT_TN noexcept
#  else
#    define CONSTEXPR11_TN constexpr
#    define CONSTEXPR14_TN constexpr
#    define NOEXCEPT_TN noexcept
#  endif
#else  // _MSC_VER
#  if _MSC_VER < 1900
#    define CONSTEXPR11_TN
#    define CONSTEXPR14_TN
#    define NOEXCEPT_TN
#  elif _MSC_VER < 2000
#    define CONSTEXPR11_TN constexpr
#    define CONSTEXPR14_TN
#    define NOEXCEPT_TN noexcept
#  else
#    define CONSTEXPR11_TN constexpr
#    define CONSTEXPR14_TN constexpr
#    define NOEXCEPT_TN noexcept
#  endif
#endif  // _MSC_VER

class static_string
{
    const char* const p_;
    const std::size_t sz_;

public:
    typedef const char* const_iterator;

    template <std::size_t N>
    CONSTEXPR11_TN static_string(const char(&a)[N]) NOEXCEPT_TN
        : p_(a)
        , sz_(N-1)
        {}

    CONSTEXPR11_TN static_string(const char* p, std::size_t N) NOEXCEPT_TN
        : p_(p)
        , sz_(N)
        {}

    CONSTEXPR11_TN const char* data() const NOEXCEPT_TN {return p_;}
    CONSTEXPR11_TN std::size_t size() const NOEXCEPT_TN {return sz_;}

    CONSTEXPR11_TN const_iterator begin() const NOEXCEPT_TN {return p_;}
    CONSTEXPR11_TN const_iterator end()   const NOEXCEPT_TN {return p_ + sz_;}

    CONSTEXPR11_TN char operator[](std::size_t n) const
    {
        return n < sz_ ? p_[n] : throw std::out_of_range("static_string");
    }
};

inline
std::ostream&
operator<<(std::ostream& os, static_string const& s)
{
    return os.write(s.data(), s.size());
}

template <class T>
CONSTEXPR14_TN
static_string
type_name()
{
#ifdef __clang__
    static_string p = __PRETTY_FUNCTION__;
    return static_string(p.data() + 31, p.size() - 31 - 1);
#elif defined(__GNUC__)
    static_string p = __PRETTY_FUNCTION__;
#  if __cplusplus < 201402
    return static_string(p.data() + 36, p.size() - 36 - 1);
#  else
    return static_string(p.data() + 46, p.size() - 46 - 1);
#  endif
#elif defined(_MSC_VER)
    static_string p = __FUNCSIG__;
    return static_string(p.data() + 38, p.size() - 38 - 7);
#endif
}

This code will auto-backoff on the constexpr if you're still stuck in ancient C++11. And if you're painting on the cave wall with C++98/03, the noexcept is sacrificed as well.

C++17 Update

In the comments below Lyberta points out that the new std::string_view can replace static_string:

template <class T>
constexpr
std::string_view
type_name()
{
    using namespace std;
#ifdef __clang__
    string_view p = __PRETTY_FUNCTION__;
    return string_view(p.data() + 34, p.size() - 34 - 1);
#elif defined(__GNUC__)
    string_view p = __PRETTY_FUNCTION__;
#  if __cplusplus < 201402
    return string_view(p.data() + 36, p.size() - 36 - 1);
#  else
    return string_view(p.data() + 49, p.find(';', 49) - 49);
#  endif
#elif defined(_MSC_VER)
    string_view p = __FUNCSIG__;
    return string_view(p.data() + 84, p.size() - 84 - 7);
#endif
}

I've updated the constants for VS thanks to the very nice detective work by Jive Dadson in the comments below.

Update:

Be sure to check out this rewrite below which eliminates the unreadable magic numbers in my latest formulation.

A

Try:

#include <typeinfo>

// …
std::cout << typeid(a).name() << '\n';