Explicit Type Conversion and Multiple Simple Type Specifiers

Question

To value initialize an object of type T, one would do something along the lines of one of the following:

T x = T();
T x((T()));

<

Answer 1

Hmm, sometimes you need a typedef. If it doesn't say a diagnostic is required, then it's not incorrect for them to support this. Nevertheless, for portability, you can use a typedef (uint16_t or uint64_t, although those might not be right), or quote the typename with a template:

iterator<void, unsigned long>::value_type( 5 )

How's that for unreasonably verbose?

Edit: Duh, or simply 5ul. That leaves unsigned short, unsigned char, and signed char as the only types you can't easily explicitly construct.

Answer 2

In §7.1.5.2, keep reading down to table 7, which has the full list of what's allowed as a simple specifier (which does include "unsigned int").

Answer 3

I posted this question to comp.lang.c++.moderated.

Daniel Krügler of the C++ standards committee agreed with the interpretation that unsigned int is a combination of simple type specifiers, and is not itself a simple type specifier.

Concerning the caption of table 7 referenced by Jerry Coffin, Krügler says:

I agree that the header of Table 7 (which is Table 9 in the most recent draft N3000) is somewhat misleading, but the preceeding text in [dcl.type.simple]/2 looks very clear to me, when it says:

Table 7 summarizes the valid combinations of simple-type-specifiers and the types they specify."

(I apologize it took me so long to post this back here from the newsgroup; it completely slipped my mind)

Answer 4

7.1.5.2:

The simple-type-specifiers specify either a previously-declared user-defined type or one of the fundamental types`

This implies that unsigned int i = unsigned int() is legal, since unsigned int is a fundamental type (and thus a simple-type-specifier, see 3.9.1).

same applies for types like:

long double
long long
long long int
unsigned long
unsigned long long int
short int
...