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## Getting a Program Right (4, Also Includes 3')

This is the fourth article about a seemingly naïve question: how do we write a binary search program? The first article appeared here.

The first two program attempts were wrong. The third article introduced a new variant reading as follows:

-- Program attempt #3.

**from**

i := 1 ; j := n

**until **i = j **loop**

m := (i + j + 1) // 2

**if **t [m] ≤ x **then**

i := m + 1

**else**** **

j := m

**end**

**end**

**if **1 ≤ i **and **i ≤ n **then Result **:= i **end
** -- If not,

The question was: is it right?

I know, you have every right to be upset at me, but the answer here too is no.

Consider a two-element array t = [0 0] (so n = 2, remember that our arrays are indexed from 1 by convention) and a search value x = 1. The successive values of the variables and expressions are:

m i j i + j + 1

After initialization: 1 2 4

i ≠ j, so enter loop: 2 3 2 6 -- First branch of "if" since t [2] < x

i ≠ j, enter loop again: 3 ⚠ -- Out-of-bounds memory access!

-- (trying to access non-existent t [3])

Oops!

Note that we could hope to get rid of the array overflow by initializing i to 0 rather than 1. This variant (version #3') is left as a bonus question to the patient reader. (*Hint*: it is also not correct. Find a counter-example.)

OK, this has to end at some point. What about the following version (#4): is it right?

-- Program attempt #4.

**from**

i := 0 ; j := n + 1

**until **i = j **loop**

m := (i + j) // 2

**if **t [m] ≤ x **then**

i := m + 1

**else**** **

j := m

**end**

**end**

**if **1 ≤ i **and **i ≤ n **then Result **:= i **end**

Answer on Friday.

*Post-publication note*: the announced fifth ("Friday") article was published here.

**Bertrand Meyer** is chief technology officer of Eiffel Software (Goleta, CA), professor and provost at the Schaffhausen Institute of Technology (Switzerland), and head of the software engineering lab at Innopolis University (Russia).

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