Adding several numbers without a loop

[super n00b]

Is there a shorter way I can add 6 numbers without using a loop?

Here is my code:

sum = 1 + 2 + 3 + 4 + 5 + 6;


Yes, this is homework, but it is acceptable to submit the assignment the long way. I just want to reduce 6 variable declarations to 1, if possible (without using a loop). The textbook doesn't show a way, so either I overlooked it or it's not possible. . .

[Edward01]

not sure if that's what you need but:

int x,sum;
while(x!=0)
{
	cin >> x;
	sum += x;
}

[TwilightSpectre]

I don't think @super n00b wants a loop. If know you have 6 numbers, you can just use an array:

#include <iostream>

int main() {
    int vals[6] = {0};
    for (int i = 0; i < 6; ++i)
        std::cin >> vals[i];

    // adding
    int sum = vals[0] + vals[1] + vals[2] + vals[3] + vals[4] + vals[5] + vals[6];
    std::cout << "Sum: " << sum << std::endl;
    return 0;
}

Of course, really a loop should be used, but considering this is an assignment, whatever. Just never do it again.

[cire]

That's what we call a loop, Edward01.

You could do so using recursion, if each number in the sequence can be derived from the last:

#include <iostream>

int sumtorial( int value )
{
    if ( value )
    {
    	std::cout << value ;
    	
    	if (value - 1)
    		std::cout << " + " ;
    		
        value += sumtorial(value-1) ;
    }
    
    return value ;
}

int main()
{
    int result = sumtorial(6) ;
    std::cout << " = " << result << '\n' ;
}

http://ideone.com/o5nBni

[long double main]

If you have the numbers in an array or a std::vector or something, you could use std::accumulate:

1 2 3 4 5 6 7 8 9 10

#include <iostream> #include <numeric> #include <iterator> int main() { int array[6] = {1,2,3,4,5,6}; std::cout << "Sum: "; std::cout << std::accumulate(std::begin(array), std::end(array), 0); }

Sum: 21

[IceThatJaw]

^^^

That's just abstracting from the loop. lol