09-06-2011, 03:30 PM
Is that really correct? If "int" on the particular system is 16 bits, then
a will be integer promoted to type int.
3 is of type int.
(a > 3) will be of type int.
b will remain unsigned int
5u will remain unsigned int
(b > 5u) will be of type unsigned int
(a > 3) && (b > 5u), the usual arithmetic conversions take place, converting the left-side expression to unsigned int.
a will be integer promoted to type int.
3 is of type int.
(a > 3) will be of type int.
b will remain unsigned int
5u will remain unsigned int
(b > 5u) will be of type unsigned int
(a > 3) && (b > 5u), the usual arithmetic conversions take place, converting the left-side expression to unsigned int.
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