为什么此 C++ 代码的 output 错误？ （hackerrank的问题之一）

Question

This is the program for printing out sum of array elements. It is showing run time error. The output is coming out to be 0 instead of printing out the sum of the elements.

#include<iostream.h>
using namespace std;
void simpleArraySum()   
{
    int ar[100],n,i,sum=0;

    for(i=0;i<n;i++)
    {
        sum=sum + ar[i];
    }

    cout<<sum;
}
int main()
{
    int ar[100],n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>ar[i];
    }

    simpleArraySum();
    return 0;
}

Answer 1

On this line in your main:

int ar[100], n;

You create an array of 100 elements. You later fill that array using cin

for(int i = 0 ; i < n ; i++)
{
    cin >> ar[i];
}

Then you do nothing with that array. You are not calculating any sum. You let that array go, forgotten.

Then, you call a simpleArraySum function. That function is creating an entirely new, distinct array .

//  v-----v------There
int ar[100],n,i,sum=0;

That array has no value assigned to it. In fact, reading from it is undefined behavior.

What you want is to receive that array in the arguments of your function:

void simpleArraySum(int* ar, int n) {
    // ...
}

And call it like that in your main:

simpleArraySum(ar, 100);

Answer 2

You can avoid the issues of arrays and functions by not using them:

int main()
{
  int quantity = 0;
  std::cin >> quantity;
  int sum = 0;
  int value;
  while (std::cin >> value)
  {
     sum += value;
  }
  std::cout << sum << "\n";
  return EXIT_SUCCESS;
}

Answer 3

In simpleArraySum , the variable n is uninitialized. So this loop:

for(i=0;i<n;i++)

invokes undefined behavior when reading from n .

Also, you are summing a different array in the function, than the one you read in mian . It seems that you need to pass in the array from main to this function:

void simpleArraySum(int *ar, int n) {

and call it like this:

simpleArraySum(ar, n);

Finally, you don't even need a function for this, since there is an existing algorithm std::accumulate that you can use:

cout << std::accumulate(ar, ar + n, 0);

Answer 4

In the function, you're adding the elements of ar which is local to the function simpleArraySum() and is not of the array ar that is local to main() .

So, pass the array and its length to the function and return its sum. Here is your corrected code:

#include<iostream>
using namespace std;
void simpleArraySum(int ar[], int n)   
{
    int i, sum = 0;

    for(i=0;i<n;i++)
    {
        sum=sum + ar[i];
    }

    cout<<sum;
}
int main()
{
    int ar[100],n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>ar[i];
    }

    simpleArraySum(ar, n);
    return 0;
}