Recursive function in c++

Question

All,

I'm writing a recursive function to do the following:

//addbig( ) -- This function is sent an array of integers and the length of the array.  
//It returns the sum of all integers in the array that are larger than 1000.

Somehow my function is not working. It is giving me zero as the output.

long addbig (const int arrInt[],int l)
{
        if (l == 0)
            return 0;
    else if(arrInt[l]>1000)
        return  arrInt[l] + addbig (arrInt,l-1);
    else
        return  addbig (arrInt,l-1);
}

My integer array is:

 int  arrInt[10]={1000,1,1000,2,1000,3,1000,4,1000,5};

Could anybody shed some light as to why this is not working, and help me a bit. haha no pun intended

@Daboyzuk is absolutely correct. 1000 is equal to 1000, not greater. Try 1001 ;) — Phil_12d3
– Phil_12d3, Commented Nov 18, 2013 at 17:09
In addition, it appears you're using l as both the index of the element you are trying to look at and the number of elements left to look at. As the code is written, you will never look at arrInt[0] — bengoesboom
– bengoesboom, Commented Nov 18, 2013 at 17:09
Why does this need to be recursive anyway? Just looping and keeping a sum would do the job...? — Phil_12d3
– Phil_12d3, Commented Nov 18, 2013 at 17:11

Zac Howland · Accepted Answer · 2013-11-18 17:12:11Z

4

First, none of the numbers in your test array are larger than 1000. So you would get 0.

Second, you are invoking UB:

long addbig (const int arrInt[],int l)
{
    if (l == 0)
        return 0;
    else if(arrInt[l] > 1000) // PROBLEM!
        return  arrInt[l] + addbig (arrInt,l-1);
    else
        return  addbig (arrInt,l-1);
}

If l is your array length, the first time you call this function will access 1 element beyond the array. What I think you want is:

long addbig (const int arrInt[],int l)
{
    if (l == 0)
        return 0;
    return addbig(arrInt, l - 1) + (arrInt[l - 1] > 1000 ? arrInt[l - 1] : 0);
}

answered Nov 18, 2013 at 17:12

Zac Howland

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Comments

RouteMapper · Accepted Answer · 2013-11-18 19:57:26Z

0

Your base case is wrong. I fixed it below. Note: none of the numbers in arrInt are greater than 1000, so you will always get zero as output.

long addbig (const int arrInt[],int l)
{
        if (l <= 0)
            return 0;
    else if(arrInt[l-1]>1000)
        return  arrInt[l-1] + addbig (arrInt,l-1);
    else
        return  addbig (arrInt,l-1);
}

edited Nov 18, 2013 at 19:57

answered Nov 18, 2013 at 17:11

RouteMapper

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3 Comments

Zac Howland Over a year ago

This will still invoke UB on the first call.

Zac Howland Over a year ago

This should not be the accepted answer as written. The else if (arrInt[l] > 1000) condition will invoke UB.

Gabriel L. Over a year ago

Wrong. What if l=0 ? arrInt[-1] is not defined.

Vlad from Moscow · Accepted Answer · 2013-11-18 17:16:04Z

0

long addbig( const int a[], size_t n )
{
   const int LIMIT = 1000;
   return ( n == 0 ? 0 : a[0] > LIMIT + addbig( a + 1, n - 1 ) );
}

answered Nov 18, 2013 at 17:16

Vlad from Moscow

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2 Comments

Zac Howland Over a year ago

This would just count the number of elements greater than the limit, not compute their sum.

Vlad from Moscow Over a year ago

@ Zac Howland you are right. I did not read the post carefully.:)

user2972135 · Accepted Answer · 2013-11-18 17:40:25Z

0

This can help you :D

#include <iostream>

using namespace std;

long addbig(const int arrInt[], int l)
{
    if (l == 0)
        return 0;
    else if (arrInt[l - 1] > 1000)
        return arrInt[l - 1] + addbig(arrInt, l - 1);
    else
        return addbig(arrInt, l - 1);
}

int main()
{
    int  arrInt[10] = {1001, 1, 2000, 2, 1000, 3, 1000, 4, 1000, 5};

    cout << addbig(arrInt, 10) << endl;

    return 0;
}

answered Nov 18, 2013 at 17:40

user2972135

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