Re: value of the constant expression 1<<(1?1:1) < 0x9999

[Peter Nilsson]

Keith Thompson wrote:
> ...
> I'll assume that types int and unsigned int are 16 bits.
>
> 1<<(1?1:1) has the value 2 and is of type int.
>
> 0x9999 has the value 39321 and is of type unsigned int.
>
> The "usual arithmetic conversions" are applied to the operands
> of "<". This converts the left operand, 2, from int to unsigned int.
>
> So the expression 1<<(1?1:1) < 0x9999 is equivalent to
> 1U < 39321U,

ITYM:  2U < 39321U

--
Peter