long long int vs. long int vs. int64_t in C++

Question

I experienced some odd behavior while using C++ type traits and have narrowed my problem down to this quirky little problem for which I will give a ton of explanation since I do not want to leave anything open for misinterpretation.

Say you have a program like so:

#include <iostream> #include <cstdint>  template <typename T> bool is_int64() { return false; }  template <> bool is_int64<int64_t>() { return true; }  int main() {  std::cout << "int:\t" << is_int64<int>() << std::endl;  std::cout << "int64_t:\t" << is_int64<int64_t>() << std::endl;  std::cout << "long int:\t" << is_int64<long int>() << std::endl;  std::cout << "long long int:\t" << is_int64<long long int>() << std::endl;   return 0; } 

In both 32-bit compile with GCC (and with 32- and 64-bit MSVC), the output of the program will be:

int:           0 int64_t:       1 long int:      0 long long int: 1 

However, the program resulting from a 64-bit GCC compile will output:

int:           0 int64_t:       1 long int:      1 long long int: 0 

This is curious, since long long int is a signed 64-bit integer and is, for all intents and purposes, identical to the long int and int64_t types, so logically, int64_t, long int and long long int would be equivalent types - the assembly generated when using these types is identical. One look at stdint.h tells me why:

# if __WORDSIZE == 64 typedef long int  int64_t; # else __extension__ typedef long long int  int64_t; # endif 

In a 64-bit compile, int64_t is long int, not a long long int (obviously).

The fix for this situation is pretty easy:

#if defined(__GNUC__) && (__WORDSIZE == 64) template <> bool is_int64<long long int>() { return true; } #endif 

But this is horribly hackish and does not scale well (actual functions of substance, uint64_t, etc). So my question is: Is there a way to tell the compiler that a long long int is the also a int64_t, just like long int is?

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My initial thoughts are that this is not possible, due to the way C/C++ type definitions work. There is not a way to specify type equivalence of the basic data types to the compiler, since that is the compiler's job (and allowing that could break a lot of things) and typedef only goes one way.

I'm also not too concerned with getting an answer here, since this is a super-duper edge case that I do not suspect anyone will ever care about when the examples are not horribly contrived (does that mean this should be community wiki?).

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Append: The reason why I'm using partial template specialization instead of an easier example like:

void go(int64_t) { }  int main() {     long long int x = 2;     go(x);     return 0; } 

is that said example will still compile, since long long int is implicitly convertible to an int64_t.

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Append: The only answer so far assumes that I want to know if a type is 64-bits. I did not want to mislead people into thinking that I care about that and probably should have provided more examples of where this problem manifests itself.

template <typename T> struct some_type_trait : boost::false_type { };  template <> struct some_type_trait<int64_t> : boost::true_type { }; 

In this example, some_type_trait<long int> will be a boost::true_type, but some_type_trait<long long int> will not be. While this makes sense in C++'s idea of types, it is not desirable.

Another example is using a qualifier like same_type (which is pretty common to use in C++0x Concepts):

template <typename T> void same_type(T, T) { }  void foo() {     long int x;     long long int y;     same_type(x, y); } 

That example fails to compile, since C++ (correctly) sees that the types are different. g++ will fail to compile with an error like: no matching function call same_type(long int&, long long int&).

I would like to stress that I understand why this is happening, but I am looking for a workaround that does not force me to repeat code all over the place.

Answer 1

You don't need to go to 64-bit to see something like this. Consider int32_t on common 32-bit platforms. It might be typedef'ed as int or as a long, but obviously only one of the two at a time. int and long are of course distinct types.

It's not hard to see that there is no workaround which makes int == int32_t == long on 32-bit systems. For the same reason, there's no way to make long == int64_t == long long on 64-bit systems.

If you could, the possible consequences would be rather painful for code that overloaded foo(int), foo(long) and foo(long long) - suddenly they'd have two definitions for the same overload?!

The correct solution is that your template code usually should not be relying on a precise type, but on the properties of that type. The whole same_type logic could still be OK for specific cases:

long foo(long x); std::tr1::disable_if(same_type(int64_t, long), int64_t)::type foo(int64_t); 

I.e., the overload foo(int64_t) is not defined when it's exactly the same as foo(long).

[edit] With C++11, we now have a standard way to write this:

long foo(long x); std::enable_if<!std::is_same<int64_t, long>::value, int64_t>::type foo(int64_t); 

[edit] Or C++20

long foo(long x); int64_t foo(int64_t) requires (!std::is_same_v<int64_t, long>); 

Answer 2

Do you want to know if a type is the same type as int64_t or do you want to know if something is 64 bits? Based on your proposed solution, I think you're asking about the latter. In that case, I would do something like

template<typename T> bool is_64bits() { return sizeof(T) * CHAR_BIT == 64; } // or >= 64