By Michael Butler, Sylvain Conchon, Fatiha Zaïdi
This ebook constitutes the refereed complaints of the seventeenth foreign convention on Formal Engineering tools, ICFEM 2015, held in Paris, France, in November 2015. The 27 revised complete papers offered have been rigorously reviewed and chosen from eighty two submissions. The papers hide quite a lot of themes within the zone of formal equipment and software program engineering and are dedicated to advancing the cutting-edge of employing formal tools in perform. They concentration particularly on combos of conceptual and methodological facets with their formal starting place and gear support.
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Additional resources for Formal Methods and Software Engineering: 17th International Conference on Formal Engineering Methods, ICFEM 2015, Paris, France, November 3-5, 2015, Proceedings
For sign correctness, the most problematic case is computing the average of 0 and 2Ei which gives 0, even if computed correctly, as rounding is to nearest, ties to even. In the other cases, the sign is correct. We will therefore prove the following properties concerning underﬂow: if the average is exactly 0, then the algorithm returns 0. If the absolute value of the average is greater or equal to 2Ei , then the returned value is non-zero. 1 The average1 Function The average1 function is the simplest one, the naive one to compute the average.
The LockOrder monitor looks surprisingly similar to the one in Fig. 1. Fig. 6. Internal shallow Scala DSL: lock order event deﬁnitions and monitor. The complete implementation of the internal DSL is less than 200 lines of code, including printing routines for error messages. Conversely, the similar external DSL is approximately 1500 lines (nearly an order of magnitude more code) and less expressive. The parts of the implementation of the internal shallow DSL directly relevant for the example in Fig.
Then we have two subcases, depending on the magnitude of y. Formal Veriﬁcation of Programs Computing the Floating-Point Average 25 If |y| ≥ 2p+Ei , then we have the same property: y 2 = y2 . Then r = (x 2) ⊕ (y 2) = x2 ⊕ y2 = ◦ x+y . 2 If |y| < 2p+Ei , then it is subnormal and the division may be inexact. But x is big enough so that this error is too small to impact the result. More precisely, . we prove that r = (x 2) ⊕ (y 2) = x2 ⊕ (y 2) = x2 = ◦ x+y 2 This is proved by using twice the following result: given a ﬂoating-point ) number f and a real h such that 2p+ei ≤ |f | and |h| ≤ ulp(f 4 , then ◦(f + h) = f .