How to deal with std::out_of_range error in the given code?

Question

Below the C++ code is supposed to reverse the vector.

#include <bits/stdc++.h>

using namespace std;

vector<int> reverseArray(vector<int> a){
    vector<int> b;
    for (int i=0;i<a.size();i++){
        b.at(i) = a.at(a.size()-i-1);
    }
    return b;
}

int main(){
    int input;
    vector<int> arr;

    // read vector
    while(cin >> input){
        arr.push_back(input);
    }

    // print vector
    for (int i=0;i<arr.size();i++){
        cout << arr.at(i) << " ";
    }
    cout << endl <<"Reversed vector is:- ";
    vector<int> r_arr = reverseArray(arr);

    // print reversed vector
    for (int i=0;i<r_arr.size();i++){
        cout << arr.at(i) << " ";
    }
}

However, an error is being thrown as follows:-

terminate called after throwing an instance of 'std::out_of_range'
  what():  vector::_M_range_check: __n (which is 0) >= this->size() (which is 0)
Aborted (core dumped)

I checked that a.size()-i-1 is varying from 5 to 0 for a.size() = 6. Why is the code not working then? Where is the problem?

Please help, I'm learning C++ STL.

t.niese · Accepted Answer · 2019-12-27 17:19:44Z

4

std::vector<int> b uses the default constructor (1) of std::vector which initializes the vector with a size of 0 so b.at(i) will always be out of range.

You need to initialize it with the same size as a using std::vector<int> b(a.size()).

But reversing a container is much easier using the reverse iterators (rbegin(), rend()):

std::vector<int> reverseArray(vector<int> a){
    std::vector<int> b(a.rbegin(), a.rend());
    return b;
}

so you actually don't need a function for that:

cout << endl << "Reversed vector is:- ";
std::vector<int> r_arr(arr.rbegin(), arr.rend());

And if it is only about printing/iterating it in revers order, then you don't even need an intermediate vector:

// print vector reversed
for (auto it=arr.rbegin(); it!=arr.rend(); it++){
  cout << (*it) << " ";
}

And if you want to reverse an existing container in place and you don't need the original order you can use std::reverse

std::reverse(std::begin(arr), std::end(arr));
for( auto i: arr ) {
  cout << i << " ";
}

Except for learning purposes, there is absolutely no reason to manually reverse an array with a loop.

edited Dec 27, 2019 at 17:19

answered Dec 27, 2019 at 16:59

t.niese

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2 Comments

t.niese Over a year ago

@user9194161 what do you mean with that? In your code? in my code?

user9194161 Over a year ago

Sorry! It was in my code. It is working correctly. Thank you sir.

Component 10 · Accepted Answer · 2019-12-27 17:01:48Z

2

You need to ensure that b is the same size as a before you start arbitrarily assigning to it. Just an extra constructor parameter should do it:

vector<int> reverseArray(vector<int> a){
    vector<int> b(a.size());
    for (int i=0;i<a.size();i++){
        b.at(i) = a.at(a.size()-i-1);
    }
    return b;
}

Also, your final loop for printing out the reversed array, is indexing the wrong array in the cout. It should read:

// print reversed vector
for (int i=0;i<r_arr.size();i++){
    cout << r_arr.at(i) << " ";
}

answered Dec 27, 2019 at 17:01

Component 10

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