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C++: long long int vs. long int vs. int64_t

Posted by: admin November 29, 2017 Leave a comment

Questions:

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?


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?).


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.


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.

Answers:

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);

Questions:
Answers:

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

Questions:
Answers:

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?

This is a good question or problem, but I suspect the answer is NO.

Also, a long int may not be a long long int.


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

I believe this is libc. I suspect you want to go deeper.

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

32-bit Linux uses the ILP32 data model. Integers, longs and pointers are 32-bit. The 64-bit type is a long long.

Microsoft documents the ranges at Data Type Ranges. The say the long long is equivalent to __int64.

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

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

64-bit Linux uses the LP64 data model. Longs are 64-bit and long long are 64-bit. As with 32-bit, Microsoft documents the ranges at Data Type Ranges and long long is still __int64.

There’s a ILP64 data model where everything is 64-bit. You have to do some extra work to get a definition for your word32 type. Also see papers like 64-Bit Programming Models: Why LP64?


But this is horribly hackish and does not scale well (actual functions of substance, uint64_t, etc)…

Yeah, it gets even better. GCC mixes and matches declarations that are supposed to take 64 bit types, so its easy to get into trouble even though you follow a particular data model. For example, the following causes a compile error and tells you to use -fpermissive:

#if __LP64__
typedef unsigned long word64;
#else
typedef unsigned long long word64;
#endif

// intel definition of rdrand64_step (http://software.intel.com/en-us/node/523864)
// extern int _rdrand64_step(unsigned __int64 *random_val);

// Try it:
word64 val;
int res = rdrand64_step(&val);

It results in:

error: invalid conversion from `word64* {aka long unsigned int*}' to `long long unsigned int*'

So, ignore LP64 and change it to:

typedef unsigned long long word64;

Then, wander over to a 64-bit ARM IoT gadget that defines LP64 and use NEON:

error: invalid conversion from `word64* {aka long long unsigned int*}' to `uint64_t*'