On fre, 2014-06-13 at 15:33 +0100, Andrew Haley wrote: > On 06/13/2014 07:42 AM, Sandeep Kumar Singh wrote: > > > I need some help/clarification, > > I am expecting zero value of 16-bit signed value 0x8000 after left shift by 2 > > It's undefined. > A small side question is implemention defined the same as implemention defined in this case? > ISO/IEC 9899:1999 (E) > > 6.5 Expressions > > 5 If an exceptional condition occurs during the evaluation of an > expression (that is, if the result is not mathematically defined or > not in the range of representable values for its type), the behavior > is undefined. > > > I have verified the same behavior with 32-bit compilers and expected (val2=0) > > with 16-bit compiler. I can correct this by shifting 18 in place of 2 as a > > workaround. I want some verify my understanding for this point and another > > way to get expected result (if any). > > Use an unsigned type. > 6.2.5 Types > > 9 ... A computation involving unsigned operands can never overflow, because a > result that cannot be represented by the resulting unsigned integer > type is reduced modulo the number that is one greater than the largest > value that can be represented by the resulting type. > > Andrew. > The result is still 0x8000 and not 0 as the original author expects. /Åke
