Incomprehensible infinite backward loop C++

Question

I am new to C++ and I am trying to achieve a backward for loop, I founded solutions which works very well but I want to know what my version is not correct and making an infinite loop.

Here is a working version that I founded (I don't understand how we can decrease i in the condition..) :

for (unsigned i = size ; i-- > 0 ; )
{
  // do stuff with i
}

Here is a version I wrote which works but don't go down to 0 (this way seems more logical to me) :

for (unsigned i = size-1 ; i > 0 ; i--)
{
  // do stuff with i
}

If I say for exemple n=10, I will get this if I print i in the loop :

And here is the version which for me is the more logical and should go down to zero but is providing an infinite loop.

for (unsigned i = size-1 ; i >= 0 ; i--)
{
  // do stuff with i
}

Could someone explain to me why the last version isn't working and what is the best choice to make ?

It's an unsigned integer. When you subtract 1 from 0 qua unsigned int, you get a very large positive number rather than -1, because -1 isn't in the range of values representable as unsigned integers. Change it to a signed int (if your upper limit is in the range representable in that data type). — Hammerite, Jun 03 '16 at 09:08
[This reference should tell you all you need to know.](http://en.cppreference.com/w/cpp/language/operator_arithmetic#Overflows) — Some programmer dude, Jun 03 '16 at 09:09
Worth noting since you start from "size": the c++11 provide `std::rbegin()` and `std::rend()`, so iterate-able containers can be iterated backwards by `for (auto it = std::rbegin(values); it != std::rend(values); ++it) { std::cout << std::distance(std::begin(values), it) << std::endl; }`, — stefaanv, Jun 03 '16 at 09:27
A good compiler of code analyser should warn that such condition is always-true — Ajay, Jun 03 '16 at 09:28

score 5 · Accepted Answer · edited May 23 '17 at 12:15

5

An unsigned number is always >= 0. (When it reaches zero a further decrement sets it to std::numeric_limits<unsigned>::max())

So your final for loop is equivalent to

for (unsigned i = size-1 ; true ; i--)

which, of course, loops forever.

In your first loop, you have i-- > 0 as the stopping condition. When i is zero, i-- is an expression with value zero (so the loop halts), despite the fact that i is then set to std::numeric_limits<unsigned>::max(). Some folk like (me included; cue the downvotes) to write i-->0 and regard --> as the slide operator. See What is the "-->" operator in C++?

edited May 23 '17 at 12:15

Community

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answered Jun 03 '16 at 09:09

Bathsheba

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Thanks ! Can you tell me more about the `i--` in the condition for the first version ? I didn't that was possible.. – oktomus Jun 03 '16 at 09:14
1

Added that. It's rather elegant, if a little hard to follow, especially at 3am when you're debugging something! Perhaps therefore worth avoiding. – Bathsheba Jun 03 '16 at 09:16
1

That's a good question! It really depends on your field and so is subjective. I'm a scientific programmer so *love* the `-->` idiom, especially when working with `unsigned` types. When I retire all my poor colleagues will be expected to understand the idiom. But I think that could be the exception rather than the norm. Another approach would be to use `signed` types, or use a different loop structure. – Bathsheba Jun 03 '16 at 09:22
Ok thanks, I think I will go for the `unsigned` types – oktomus Jun 03 '16 at 09:47

score 2 · Answer 2 · answered Jun 03 '16 at 09:10

The statement i >= 0 is always true because i is unsigned which means that is never below zero. If you decrease the value while i is zero, there will occur a so-called underflow and it will have a very high number.

The first version certainly gets the job done, so I would stick to it.

Incomprehensible infinite backward loop C++

2 Answers2