CODEWITHSUNDEEP
c++debuggingstldeque
Computernerd
Computernerd

Reputation: 7766

simple deque programme exhibit weird behaviour

I am using a deque to store integers from 1 to 10 and output to console 1 to 10 but for some unknown reason it outputs 11 for every loop . I cannot figure out what i am doing wrong. 

#include <iostream>
#include <cstring>
#include <deque>
using namespace std;




int main()
{
    deque <int> deq;

    int i;

    for ( i= 1 ;i <=10 ;i++)
    {
       deq.push_front(i);
    }

    deque <int>::iterator d2;

    d2 = deq.begin();

    while (d2 != deq.end() )
    {
        cout<<i
            <<endl;

            d2++;
    }


}

Thanks for the help , i understand the problem already

Upvotes: 0

Views: 102

Answers (5)

iqbal
iqbal

Reputation: 35

If you print i inside the while loop it produce always 11 because above your code. You already use int i in the for loop, with the last value is 10. Debug your code with ddebugging tools ini your editor again, you will know what you'r doing wrong.

while (d2 != deq.end() )
{
cout<<i<<endl;
d2++;
}

The above while always produce 11 every iteration.

Upvotes: 0

Glenn Teitelbaum
Glenn Teitelbaum

Reputation: 10333

int main()
{
    deque <int> deq;

    int i;

    for ( i= 1 ;i <=10 ;i++)
    {
       deq.push_front(i);
    }

===> i is 11 here (i <= 10) is false at 11

    deque <int>::iterator d2;

    d2 = deq.begin();

===> here you really want to use d2

    while (d2 != deq.end() )
    {
        cout<<i

===> here you print i instead of d2

             <<endl;


            d2++;
    }


}

Upvotes: 1

Chad
Chad

Reputation: 19032

You need to print out the value contained by the iterator, not i:

while (d2 != deq.end() )
{
    // wrong!
    //cout<<i
    //    <<endl;

    cout << *d2 << endl;

        d2++;
}

As a side note, this illustrates why you should always limit the scope of variables to the smallest possible scope. Had your original loop been declared as:

for(int i = 1; i <= 10; ++i)

Then attempting to print out i (incorrectly) later would have been a compile time error.

Upvotes: 2

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

You should replace this

cout << i << endl;

with something that actually remove things from the dequeue:

cout << deq.pop_front() << endl;

To avoid simple errors like this in the future, restrict the scope of loop index variables by declaring them inside the loop header:

for (int i= 1 ;i <=10 ;i++)
{
   deq.push_front(i);
}

This way you would get a compile error trying to reference i outside the loop.

Upvotes: 1

Maria Ines Parnisari
Maria Ines Parnisari

Reputation: 17486

You're printing i, which was assigned to i=11 in the for loop.

Upvotes: 1

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