Auto failed for vector size

Question

I am trying use auto to infer the type.

for (auto i = remaining_group.size() - 1; i >= 0; --i) {
    std::cout << i;
}

I get very large number like 18446744073709534800 which is not expected. When I change auto to int it is the number I expected between 0 and 39.

Is there any reason auto will fail here?

remaining_group's type is a std::vector<lidar_point> and lidar_point is struct like:

struct LidarPoint {
  float x;
  float y;
  float z;
  uint8_t field1;
  double field2;
  uint8_t field3;
  uint16_t field4;
}

How is `remainug_group` defined? Also - maybe the size is 0? — einpoklum, May 26 '18 at 11:55
If `size` is unsigned then condition `i >= 0` is always true and you eventually get an integer underflow. — user7860670, May 26 '18 at 11:57
Possible duplicate of [Using auto in loops c++](https://stackoverflow.com/questions/17505164/using-auto-in-loops-c) — , May 26 '18 at 11:58
It only takes 2 more chars to write `size_t` instead of `auto` but the explicitness it brings is well worth it — Killzone Kid, May 26 '18 at 12:08
What compiler are you using that didn't give a warning for your code? — Marc Glisse, May 26 '18 at 12:14

songyuanyao · Accepted Answer · 2018-05-26T12:07:20.543

2

When using auto the type of i would be std::vector::size_type, which is an unsigned integer type. That means the condition i >= 0; would be always true, and if overflow happens you'll get some large numbers.

Unsigned integer arithmetic is always performed modulo 2ⁿ where n is the number of bits in that particular integer. E.g. for unsigned int, adding one to UINT_MAX gives 0, and subtracting one from 0 gives UINT_MAX.

edited May 26 '18 at 12:07

answered May 26 '18 at 11:59

songyuanyao

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Thanks! This is the right answer. Really need to be careful using auto in this case! – Alex May 26 '18 at 12:12
@songyuanyao • the static cast should not including the `- 1` part. Consider what happens when remaining_group is empty, so has size 0. – Eljay May 26 '18 at 12:58
@Eljay Correct. – songyuanyao May 26 '18 at 13:00
@Alex Here's a wordy workaround if you stick to `auto`, `auto i = static_cast(remaining_group.size()) - 1`. – songyuanyao May 26 '18 at 13:00
@Swift-FridayPie I didn't get it; the type of `i` is still an unsigned integer type? [LIVE](https://wandbox.org/permlink/YoPj3VD3A0Mu35eK) – songyuanyao May 26 '18 at 13:04
@songyuanyao hm, interesting, I found compiler bug – Swift - Friday Pie May 26 '18 at 13:08

Swift - Friday Pie · Answer 2 · 2018-05-26T12:09:25.490

1

Simple reproduction of problem:

#include <iostream>

size_t size()  {return 1;}

int main() {
   for (auto i = size() - 1; i >= 0; --i) {
    std::cout << i << std::endl;
   }
}

size() got type size_t and literal constant 1 would be promoted to size_t, in result auto would become size_t, which cannot be less than zero, resulting in infinite loop and underflow of i.

edited May 26 '18 at 12:09

answered May 26 '18 at 12:01

Swift - Friday Pie

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Oh, man, integer promotion. I would have expected the final type so be signed, but ... promotion rules. – Vorac May 26 '18 at 12:58

score 1 · Answer 3 · answered May 26 '18 at 12:39

If you need a reverse index loop, use operator -->

When you write a normal index loop, you write it with 0, size, and <.

When you write a normal reverse index loop, things become a bit wonky: you need size - 1, >=, 0, and you can't use unsigned index, because unsigned i is always >= 0 so your check i >= 0 always returns true, your loop could run forever.

With fake operator "goes to", you can use 0, size, and > to write the reverse index loop, and it doesn't matter if i is signed or unsigned:

for (auto i = a.size(); i --> 0; ) //or just i--, or i --> 1, i --> 10...
    std::cout << i << ' ';

Auto failed for vector size

3 Answers3