问题
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.
回答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);
回答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
回答3:
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*'
来源：https://stackoverflow.com/questions/4160945/long-long-int-vs-long-int-vs-int64-t-in-c