Function call misunderstanding

Question

For example, I have a recursive function which calculate n!.

 int fact(int n){
    if(n==1)
       return 1;
    return n*fact(n-1);
}

My question is why I can't replace n-1 with n--? Or can I replace it with --n?

@HolyBlackCat: `n` is used twice in the return expression, careful with order of evaluation guarantees... — Mat, Jul 02 '18 at 08:38
please pick a language : C++ != C (especially if you get into the nooks and crannies like sequencing) — Sander De Dycker, Jul 02 '18 at 08:47
@HolyBlackCat: There is a difference between n-1 and --n in this code. You will get incorrect result with --n. — , Jul 02 '18 at 08:47

score 4 · Answer 1 · answered Jul 02 '18 at 08:50

There are two reasons.

One, n-- is post-decrement. The value used as the arugment would be the original (non-decremented) value of n.

Two, n is used twice in the expression (once as the left-hand side argument of *, and once within the argument to the recursive call). Neither C nor C++ specify evaluation order in this case, so it would be unclear whether the n is evaluated before or after n-- (or even --n).

score 3 · Accepted Answer · answered Jul 02 '18 at 08:39

3

because it changes the value of n. Since the value is used to compute the expression n * fact(...), the value of n must be read after the fact is evaluated. Since --n (or n--, it does not matter here) must be passed as value to fact, the decrement must occur before the fact is invoked. Therefore the expression result will be incorrect.

answered Jul 02 '18 at 08:39

Remus Rusanu

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I would rather say "can be incorrect", there is also a chance that it will evaluate as the OP expects (with `--n` at least) – Ctx Jul 02 '18 at 08:44
It depends what value you pass to fact in case of --n. For 0, it will crash, for 1 it will be correct. And then it will be incorrect. – Jul 02 '18 at 08:49
It doesn't have to crash, since a tail recursion could be optimized to an iteration – Ctx Jul 02 '18 at 08:50
1

@JamesBond And it would yield correct results if n is evaluated, then decremented and finally fed to funct, which would be perfectly possible afaics – Ctx Jul 02 '18 at 08:51
Comments are correct, but I think OP's question is at a more superficial understanding level. – Remus Rusanu Jul 02 '18 at 08:51
1

@RemusRusanu Maybe, but this Q/A is not only for the OP, but for all interested readers – Ctx Jul 02 '18 at 08:57

Function call misunderstanding

2 Answers2