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  lt iostream gt   include  lt cstdint gt   template  lt typename T gt  bool is int64     return false     template  lt  gt  bool is int64 lt int64 t gt      return true     int main      std  cout  lt  lt   int  t   lt  lt  is int64 lt int gt     lt  lt  std  endl   std  cout  lt  lt   int64 t  t   lt  lt  is int64 lt int64 t gt     lt  lt  std  endl   std  cout  lt  lt   long int  t   lt  lt  is int64 lt long int gt     lt  lt  std  endl   std  cout  lt  lt   long long int  t   lt  lt  is int64 lt long long int gt     lt  lt  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     amp  amp     WORDSIZE    64  template  lt  gt  bool is int64 lt long long int gt      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  lt typename T gt  struct some type trait   boost  false type       template  lt  gt  struct some type trait lt int64 t gt    boost  true type        In this example  some type trait lt long int gt  will be a boost  true type  but some type trait lt long long int gt  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  lt typename T gt  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 amp   long long int amp     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

User · Accepted Answer

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 lt  std  is same lt int64 t  long gt   value  int64 t gt   type foo int64 t     edit  Or C  20 long foo long x   int64 t foo int64 t  requires   std  is same v lt int64 t  long gt

User · Answer

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  amp 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

User · Answer

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 lt typename T gt  bool is 64bits     return sizeof T    CHAR BIT    64       or  gt   64

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