16 byte integers

Question

I am compiling a C ++ program using 64 bit GCC - machine / OS / (with the -m64 option passed to g++ ). As expected, sizeof(long double) == 16 - I wonder if there is a 16-byte standard type for integers?

PS __int128_t is an artificial extension that emulates the standard type, as I understand it. Other than that, I did not find anything.

Answer 1

No, only guaranteed sizes for char , unsigned char and signed char , and they are 1 :

5.3.3 Size [expr.sizeof]

1 The sizeof operator gives the number of bytes in an object representing its operand. An operand is either an expression that is not evaluated, or a type identifier in brackets. The size operator does not apply to an expression that has a function or an incomplete or enumeration type before all its counters were declared either in brackets of the name of such types, or the value l, which denotes a bit field. sizeof (char), sizeof (signed char) and sizeof (unsigned char) are equal to 1. The result of sizeof applies to any other fundamental type (3.9.1) is determined by the implementation. [Note: in particular, sizeof (bool) and sizeof (wchar_t) are a specific implementation of .74) -end note] [Note: see 1.7 for the definition of a byte and 3.9 for the definition of an object representation. -end note]

(my emphasis)

Answer 2

The C ++ standard does not standardize the size of each type (accept char), but the difference in size relative to each other.

 A char has to be 1 byte. A short has to be >= to a char. An int has to be >= to a short. A long has to be >= to an int. and a long long has to be >= to a long. 

the same applies to float types.

I believe most compilers now support up to 8 byte integers

however, I know that some, if not most, do not have 16-byte integers.

I also think that there are some classes of objects / structures that people have developed to emulate a 16-byte integer that will work on most compilers.