# C++ Program to Add n Numbers

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In this article, you will learn and get code on adding of n numbers (or finding sum of n numbers) given by user at run-time using C++ program. For example, if user enters the value of n as 5. Then the program further asks to enter any 5 numbers. So that it can add all the 5 entered numbers and print its summation result as output.

Here are the list of approaches that are used to do the task of adding n numbers:

• Find sum of n numbers using for loop
• using while loop
• using array
• using user-defined function

At last of the article, you will also see a program that adds n natural numbers.

To add n numbers in C++ programming, you have to ask from user to enter the value of n (i.e., how many numbers he/she wants to enter), then ask to enter n numbers to perform the addition of all the given numbers and finally display the result on the screen as shown here in the following program.

## Find Sum of n Numbers using for Loop

This program is created using for loop to do the job. The question is, write a program in C++ to find and print the sum of n numbers using for loop. And here is the answer to this question:

```#include<iostream>
using namespace std;
int main()
{
int i, n, num, sum=0;
cout<<"How many numbers you want to enter ? ";
cin>>n;
cout<<"Enter "<<n<<" numbers: ";
for(i=0; i<n; i++)
{
cin>>num;
sum = sum+num;
}
cout<<"\nSum of all "<<n<<" numbers is "<<sum;
cout<<endl;
return 0;
}```

This program was build and run under Code::Blocks IDE. Here is its sample run:

Now supply or enter the value of n (that how many numbers you wants to enter and add them up) say 5. And then enter any 5 numbers. Press `ENTER` key to see the following output, that will be the summation of all the 5 numbers entered by user at run-time:

The program is very simple. That is, received the value of n first, and then created a for loop that runs n number of times to get all the numbers and add them up one by one. Therefore, if user enters the value of n as 5, then the dry run of above program goes like:

• Inside for loop, initially 0 gets initialized to i and checks the condition, that is whether the value of i is less than n's value or not
• Condition evaluates to be true, therefore program flow goes inside the loop, and receives a number inside a variable say num
• And sum+num gets initialized to sum every time after receiving the number
• Because, the initial value of sum is 0, therefore at first execution of for loop, suppose user enters first number as 10. Therefore, 0+10 or 10 gets initialized to sum. Now sum=10
• Program flow goes to update part, and increments the value of i and again checks the condition
• The condition again evaluates to be true for second time, and program flow again goes inside the loop, and executes the two statements. That is, receives a number say 20 and initializes sum+20 or 10+20 as the new value of sum. Therefore, now sum=30
• In similar way, process the operation
• The loop executes terminates, when its condition evaluated to be false
• After exiting from the loop, we'll have a variable sum that holds the summation value of all the n numbers entered by user
• Therefore, simply prints its value as output

### What if Number(s) contains Decimal ?

To handle with real numbers (numbers having decimal part), use the float data type instead of int. Therefore just replace the following statement as given in above program:

`int i, n, num, sum=0;`

with the statement given below:

`float i, n, num, sum=0;`

Rest of the things will be same.

## Sum of n Numbers using while Loop

The question is, write a program in C++ that find and prints the sum of n given numbers using while loop. The answer to this question is given below:

```#include<iostream>
using namespace std;
int main()
{
int n, i=0, num, sum=0;
cout<<"Enter the value of n: ";
cin>>n;
cout<<"Enter "<<n<<" numbers: ";
while(i<n)
{
cin>>num;
sum = sum+num;
i++;
}
cout<<"\nSum = "<<sum;
cout<<endl;
return 0;
}```

The output produced by above program looks like:

Unlike for loop, in while loop we have to initialize the loop variable before the start of while loop. As it only contains the condition statement, therefore we also have to place the updatation part inside the loop's body. Therefore, the value of i gets initialized before the loop and its value gets updated using the last statement of its body.

## Add n Numbers using Array

This program uses array to do the same job as of previous program.

```#include<iostream>
using namespace std;
int main()
{
int n, i, arr, sum=0;
cout<<"Enter the value of n (max. 50): ";
cin>>n;
cout<<"Enter "<<n<<" numbers: ";
for(i=0; i<n; i++)
{
cin>>arr[i];
sum = sum+arr[i];
}
cout<<"\nSum = "<<sum;
cout<<endl;
return 0;
}```

Here is its sample run with same input as of previous output:

All the numbers gets stored inside the array in a way that:

• First number gets stored at arr
• Second number at arr
• Third at arr
• and so on

## Add n Numbers using Function

This is the last program on adding n numbers. This program is created using a user-defined function named findSum() to do the same job. This function receives array and its size as its two argument or parameters. The array are the list of numbers, and size is the value of n. Further it finds summation of all numbers in the list and returns it. That is, its return value gets initialized to sum inside the main() function:

```#include<iostream>
using namespace std;
int findSum(int [], int);
int main()
{
int n, i, arr, sum;
cout<<"Enter the value of n (max. 50): ";
cin>>n;
cout<<"Enter "<<n<<" numbers: ";
for(i=0; i<n; i++)
cin>>arr[i];
sum = findSum(arr, n);
cout<<"\nSum = "<<sum;
cout<<endl;
return 0;
}
int findSum(int arr[], int n)
{
int i, sum=0;
for(i=0; i<n; i++)
sum = sum+arr[i];
return sum;
}```

This will produce the same output as of previous one.

## Find Sum of n Natural Numbers in C++

This program adds first n natural number. The value of n must be entered by user. For example, if the value of n gets entered by user as 5. Then the following program will find out the sum of first 5 natural numbers. That will be 1, 2, 3, 4, 5. Therefore, its summation will be 1+2+3+4+5 equals 15

```#include<iostream>
using namespace std;
int main()
{
int i, n, sum=0;
cout<<"Enter the value of n: ";
cin>>n;
for(i=1; i<=n; i++)
sum = sum+i;
cout<<"\nSum of first "<<n<<" natural numbers = "<<sum;
cout<<endl;
return 0;
}```

The sample run of above program with user input as 10 will looks like:

#### Same Program in Other Languages

C++ Online Test

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