Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Assume that t,a,b are all double (IEEE Std 754) variables, and both values of a, b are NOT NaN (but may be Inf).
After t = a - b, do I necessarily have a == b + t?
Absolutely not. One obvious case is
a=DBL_MAX,b=-DBL_MAX. Thent=INFINITY, sob+tis alsoINFINITY.What may be more surprising is that there are cases where this happens without any overflow. Basically, they’re all of the form where
a-bis inexact. For example, ifaisDBL_EPSILON/4andbis-1,a-bis 1 (assuming default rounding mode), anda-b+bis then 0.The reason I mention this second example is that this is the canonical way of forcing rounding to a particular precision in IEEE arithmetic. For instance, if you have a number in the range [0,1) and want to force rounding it to 4 bits of precision, you would add and then subtract
0x1p49.