Question about page 307-308: Chapter 4.10, Exercises 4-5

[MetallicCrimson]

Ex. 4 has a solution in the Appendix (obviously the same as here), and ex. 5 follows directly.
However, I think the book is giving inconsistent information.

Consider this section from a bit earlier:

Here, the trace table is collecting the values of x and i after a whole iteration has been completed - and because a for loop is just syntactic sugar for a while loop (which includes i := i + 1), technically the value of i is already one more. (And it ends with i = 5, which makes sense, since that's the point the execution will break out of the loop.)

This behavior is not reflected in the Appendix solution, which lists the values of i as they are at the start of each iteration. This conflicts the fact about the syntactic sugar. Sure, it matches the iteration number, but that can be (and is) different than the value of i.

So going by Example 4.10.3, the correct trace table for Ex. 4 would be:

i│1│2│3 │4 
─┼─┼─┼──┼──
a│2│7│22│67

(And similarly for ex. 5 too.)

This doesn't matter much, since (ideally) an iterator variable in a for loop isn't used for anything else, but technically makes the (book's) solution incorrect. (The fact that the example has the i line at the bottom and the exercise has it at the top shouldn't modify the values of the table.)

This is not listed in the Errata. Do you think it's an issue, or is it fine to leave it this way? Or is my reasoning incorrect?

[spamegg1]

Yes you are right, I think it should be in the Errata.
Possibly she just treated the i/k variables as iteration number. (Because then it makes sense)

I'll fix these where the table has an additional row with iteration number to clarify everything.