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

For example:

int a = 12; cout << typeof(a) << endl; 

Expected output:

int 
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Jorge Ferreira Avatar asked Sep 17 '08 10:09

Jorge Ferreira


People also ask

Can you print a variable's type in standard 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.

Can you get the type of a variable in C?

C is statically typed language. You can't declare a function which operate on type A or type B, and you can't declare variable which hold type A or type B. Every variable has an explicitly declared and unchangeable type, and you supposed to use this knowledge.

How do you print a variable in C?

This number is stored in the number variable. printf("Enter an integer: "); scanf("%d", &number); Finally, the value stored in number is displayed on the screen using printf() . printf("You entered: %d", number);

How do you get the data type of a variable in C++?

To get the datatype of variable, use typeid(x). name() of typeinfo library. It returns the type name of the variable as a string.


2 Answers

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

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.

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Howard Hinnant Avatar answered Oct 02 '22 20:10

Howard Hinnant


Try:

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

You might have to activate RTTI in your compiler options for this to work. Additionally, the output of this depends on the compiler. It might be a raw type name or a name mangling symbol or anything in between.

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Konrad Rudolph Avatar answered Oct 02 '22 21:10

Konrad Rudolph