Question

how to write program that takes a number as its argument and return the sum of 1+2+.. up to argument?

I can’t get the codes right. Can somebody help me?

#include<stdio.h>
#include<stdlib.h>                                                               
int main(int argc, char*argv[])
{
int i;
int sum =0;

if(argc !=2){
 printf("usage: %s <count> \n", argv[0]);
 exit(n);
 }

for(i=1; i<=atoi(argv[1]); i++){
 sum+=i;
}

Answer 1

First, to answer your question.

You need to actually output the result. Something like:

printf("%d\n", sum);

Or return it to whoever called the program, although that is a little unusual:

int main( int argc, char **argv ) {
    ...
    return sum;
}

But I am providing my own answer here because there is a good reason to consider doing this in a loop… At least until you’ve thought about it a bit more.

Namely, the formula (n * (n+1)) / 2 will overflow 32-bit integers and produce the wrong answer when n becomes 65536 or greater. But the 32-bit integer can itself store a sum up to n <= 92681. That means the formula by itself produces the wrong answer for roughly 30% of the solution space.

So you might think you need to loop, but there’s a little trick here. Because the formula uses both n and n+1, you can guarantee that one of those numbers is evenly divisible by 2. And therefore you can do it like this:

unsigned long n;
unsigned long sum;

n = atoi(argv[1]);

if( n == 0 || n > 92681 ) {
    printf( "The supplied value (%u) is out of range\n", n );
} else {
    if( (n % 2) == 0 ) {
        sum = (n / 2) * (n+1);
    } else {
        sum = n * ((n+1) / 2);
    }
    printf( "Sum from 1 to %u is %u\n", n, sum );
}

Now you have a simple formula that produces the same answer as the loop, at least for all values of n that don’t lead to overflowing the sum.