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Re: signed/unsigned right shift
- From: Christian Groessler <cpg at aladdin dot de>
- To: Ian Lance Taylor <ian at wasabisystems dot com>
- Cc: GCC List <gcc at gcc dot gnu dot org>
- Date: 15 Mar 2004 23:02:11 +0100
- Subject: Re: signed/unsigned right shift
On 03/14/2004 01:23:18 PM EST Ian Lance Taylor wrote:
>
>Christian Groessler <cpg@aladdin.de> writes:
>
>> >Correct, the multiplication is done as signed int. And it is
>> >implementation
>> >defined whether right shift of a negative signed int does sign fill or
>> >not.
>> >So both compilers are right; it is only your code that's wrong.
>> >
>> >C89 references: 3.3.7, 3.2.1.5, 3.2.1.1.
>>
>> Hmm, unfortunately I don't have the c89 standard document. Can
>> somebody point me to the equvalent sections in c99?
>
>Integer promotions: 6.3.1.1.
Sentence 2 says "If an int can represent all values of the original
type, the value is converted to an int; otherwise, it is converted to
an unsigned int."
Seems to suggest that the conversion "unsigned short" to "signed int"
is ok.
But sentence 3 says "The integer promotions preserve value including
sign."
So shouldn't the conversion be "unsigned short" to "unsigned int"?
regards,
chris