Don't understand what array-1 means

Question

I have a problem with understanding this code, especially the "vet-1" part. What does it mean? What item of the array does it return?

I found this exercise while practicing for an exam in my first year of University and I wondered if it has a solution. It is not my own code. Thanks!

#include <stdlib>
#include <iostream>

using namespace std;

void change(int m, int n[7]);

int main(){
    int vet[] = {1, 2, 3, 4, 5};
    change(vet[4], vet-1);
    change(0, &vet[4]);
    int i = 0;
    for (i=0; i<5; i++) cout << vet[i];
    return 0;
}

void change(int m, int n[7]) {
    (*(n+m))--; m++; n--;
}

I think it is UB as `vet-1` is out of bound array (even if not deferenced). — Jarod42, Jun 07 '18 at 09:50
@Jarod42: Yes your're correct. The only exception would be `vet + 5`. — Bathsheba, Jun 07 '18 at 09:51
It probably tries to use the fact that with C arrays [`a[b]` is the same as `*(a + b)`](https://stackoverflow.com/questions/381542/with-arrays-why-is-it-the-case-that-a5-5a), but fails by going out of bounds. — Bo Persson, Jun 07 '18 at 09:54
In general, subtracting one from a pointer is analogous to adding one to a pointer: it moves to the element just before. If the pointer points somewhere within an array, this can be perfectly legal and perfectly meaningful; it's a problem only if the pointer is already pointing at the beginning of the array. In the posted code fragment, however, `v` does point to the beginning of the array (it *is* the array!), so it's undefined. The code has other problems, and should be thrown away, not learned from. Any textbook or course that presented this code for instruction is suspect. — Steve Summit, Jun 07 '18 at 09:59
I'm assuming this code is from some book/online tutorial/teacher/school/.... . This code is a good example how to "NOT TEACH" programing in C or C++. I recommend change source. — Marek R, Nov 08 '22 at 11:47
@MarekR Yes it was 4 years ago so I don't remember exactly from which resource I found it, but it was my first year of University and this was an exercise I found somewhere. Don't know why lately this question is getting popular and getting downvoted :D — Giulio Mattolin, Nov 08 '22 at 15:15
@GiulioMattolin someone edited your question so it was shown to me. Didn't notice that it is old. — Marek R, Nov 08 '22 at 15:19

score 13 · Accepted Answer · answered Jun 07 '18 at 09:50

13

vet - 1 is an attempt at referring to the pointer to the element just before vet[0].

Actually the behaviour on doing that is undefined. So the entire program is undefined.

Nothing to understand here; move on!

answered Jun 07 '18 at 09:50

Bathsheba

231,907
34
361
483

4

_"move on!"_ quite literally. Especially if it was an interview question. – YSC Jun 07 '18 at 09:52
I thought the same, but why the program print something instead of giving a run-time error? – Giulio Mattolin Jun 07 '18 at 13:10
1

@FrancescoMarzetta: Because when you use `[]` to access an element, C++ assumes you know what you're doing. If you want an exception to be thrown, then use `.at`. – Bathsheba Jun 07 '18 at 13:12
2

@FrancescoMarzetta That's part of undefined behavior. The program *might* crash, but it doesn't have to. – dbush Jun 07 '18 at 13:12

Don't understand what array-1 means

1 Answers1