That's true, my intention was to make the value clearer. Simply writing
UINT16_MAX is easier than "(uint16_t) -1" which equals "-1 + (UINT16_MAX
+ 1)" as you also mentioned.
On 12/07/12 14:46, Ben Pfaff wrote:
On Thu, Jul 12, 2012 at 01:31:49PM +0100, Zoltan Kiss wrote:
The value represented by the macro would stay exactly the same.
See this article about this topic:
http://embeddedgurus.com/barr-code/2011/06/is-uint16_t-1-portable-c-code/
This article is wrong. (uint16_t) -1 is guaranteed to have the value
65535, because C99 defines conversion between signed and unsigned types
this way:
6.3.1.3 Signed and unsigned integers
...
Otherwise, if the new type is unsigned, the value is converted by
repeatedly adding or subtracting one more than the maximum value
that can be represented in the new type until the value is in the
range of the new type.49)
49) The rules describe arithmetic on the mathematical value, not
the value of a given type of expression.
Thus, (uint16_t) -1 has the value -1 + (UINT16_MAX + 1), that is,
UINT16_MAX.
I'm not going to take any action based on the advice of an article that
is so badly researched.
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