int64_t == long long

[John Love-Jensen]

On 7/1/08 2:17 PM, "Yang Zhang" <yanghatespam@gmail.com> wrote:

> Hi, why isn't int64_t == long long at least on 64-bit x86 Linux?

Because int64_t should be 64-bit, but long long could be 64-bit or larger.

#include <stdint.h> // from C99
#include <climits>

cout << (sizeof(int64_t) * CHAR_BIT) << endl;
cout << (sizeof(long long) * CHAR_BIT) << endl;

You can also do this:

#include <stdint.h> // from C99
#include <typeinfo>

cout << typeid(int64_t).name() << endl;
cout << typeid(long long).name() << endl;

> How do I tell what type this actually is?

typeid

> And are literals ending with LL always long long?

Yes, that's what the LL suffix means.

So the literal integer numerics are:
'A'
L'A'
65
65L
65LL
65U
65UL
65ULL

Keep in mind the portability issues surrounding use of long long, LL, and
ULL.  You may want to use the <stdint.h> INT64_C and UINT64_C macros to
construct your numeric literals.

INT64_C(65)
UINT64_C(65)

HTH,
--Eljay