Issue with Auto keyword in C++

Question

I have written a code to rotate an array of integers cyclically. For e.g. Given array - 1 2 3 4 5 6 7 8 Output array- 8 1 2 3 4 5 6 7

The problem what i am facing is, if i use auto keyword in some point then it gives weird result what is beyond my understanding. Would anyone please help me in summarizing the problem ?

#include<iostream>
#include<bits/stdc++.h>
#include <typeinfo>
using namespace std;

void rotate_one_by_one(array<int, 8> &arr)
{
    auto temp = arr[arr.size() - 1];
    //auto i = arr.size() - 2; // Output : 1 2 3 4 5 6 7 8 
    int i = arr.size() - 2;    // Output : 8 1 2 3 4 5 6 7

    for(; i > -1; --i)
    {
        arr[i+1] = arr[i];
    }
    arr.at(i+1) = temp;
}

void cyc_rotate(array<int, 8> &arr)
{
    rotate_one_by_one(arr);

    cout<<"After cyclic rotate\n";
    for(auto n : arr)
        cout<<n<<" ";
}

int main()
{
    array<int, 8> arr = {1,2,3,4,5,6,7,8};

    cyc_rotate(arr);
    return 0;
}

Please find the commented line in the code where i have mentioned the output for using auto and int type. If anyone still not able to get what i am trying to convey then please let me know.

Unrelated to your problem, but please read [Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — Some programmer dude, Jun 12 '19 at 10:54
Just a quick guess, but `auto` probably infers that the result is unsigned, unlike `int`. — rjg, Jun 12 '19 at 10:55
auto will default `i` to `std::size_t` but you defined `i` as an `int`, which is a signed number. — Neijwiert, Jun 12 '19 at 10:55
`arr.size()` is an unsigned integer. `arr.size() - 2` is unsigned. `i > -1` is equivalent to `i > 4294967295` which is always false. So the array doesn't change. — L. F., Jun 12 '19 at 10:55
Also, [why is `using namespace std;` considered bad practice?](https://stackoverflow.com/q/1452721/9716597) — L. F., Jun 12 '19 at 10:57
Turn on [compiler warnings](https://gcc.godbolt.org/z/dGp3nD) — Ayxan Haqverdili, Jun 12 '19 at 11:07

KamilCuk · Answer 1 · 2019-06-12T11:06:41.323

4

The return value of arr.size() function has unsigned type std::size_t. If variable is auto i, it becomes std::size_t type too. As it is unsigned, it can never reach a value below 0.

for(; i > -1; --i)

-1 is converted here to an unsigned value, most probably to std::numeric_limits<size_t>::max() or SIZE_MAX. As, in such case, i will never be greater then this value, the loop will never enter.

edited Jun 12 '19 at 11:06

answered Jun 12 '19 at 10:56

KamilCuk

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"the loop will loop forever" then how is OP getting an output? – Gaurav Singh Jun 12 '19 at 11:00
@GauravSingh The behaviour of an unconditionally infinite loop is undefined (unless the loop does certain things, which this loop doesn't). – eerorika Jun 12 '19 at 11:28
@Kamil Cuk It does not enter into the loop as well with auto keyword. – chinmaya Jun 12 '19 at 12:04
@chinmaya When `int` is used it _does_ enter the loop. When `auto` is used it doesn't, just as Kamil wrote in his edit before you commented. – Ted Lyngmo Jun 12 '19 at 12:55
Yup @TedLyngmo i got that later and i appreciate for those who responded to my query. Thanks a lot. – chinmaya Jun 13 '19 at 04:59

score 3 · Accepted Answer · answered Jun 12 '19 at 12:07

auto i = arr.size() - 2 makes i a variable of type std::size_t which is an unsigned integer. It can never be negative. The comparison with -1 promotes -1 to size_t by adding std::numeric_limits<std::size_t>::max() + 1 to it. Your i can't ever be greater than that so the loop will never be entered.

You could rescue your condition with for(; i != static_cast<decltype(i)>(-1); --i) or better, use std::rotate:

#include <algorithm> // std::rotate
#include <array>
#include <iostream>

int main() {
    std::array<int, 8> arr = {1, 2, 3, 4, 5, 6, 7, 8};

    std::rotate(arr.begin(), std::prev(arr.end()), arr.end());

    for(auto i : arr) std::cout << i << " ";
    std::cout << "\n";
}

Do not use the header <bits/stdc++.h>. It's non-standard and you'll never know if you get what you need, but you will for sure include much else that you don't need.

Issue with Auto keyword in C++

2 Answers2