Question

Write a for loop that assigns summedvalue with the sum of all odd values from 1 to usernum. assume usernum is always greater than or equal to 1.

Answer 1

Firstly, define a variable `usernum` and let's use 10 as an example. Then, define another variable `summedvalue` and set it as 0 initially. This variable will hold the sum of all odd numbers up to `usernum`.

Next, we start a for loop. This loop will go through each number in the range from 1 to `usernum` inclusive. That's why in the range function we provide `usernum + 1` as the second parameter.

Now, for every number `i` in this range, we check whether it is odd. A number is odd if it gives a remainder of 1 when divided by 2. In other words, we're checking if `i % 2` is not equal to 0 (`i % 2 != 0`).

If the current number `i` is indeed odd, we add it to `summedvalue` (`summedvalue += i`). Then, the loop goes to the next number and the process repeats itself until we've considered all numbers from 1 to `usernum`.

Finally, when the loop ends, the value of `summedvalue` is the sum of all odd numbers from 1 up to (and including) `usernum`. Return this `summedvalue`.

In the case of `usernum = 10`, the odd numbers between 1 to `usernum` are 1, 3, 5, 7, and 9. When you add them together, you get 25. That's why in this case, the result will be 25.

Answer 2

// Input value is usernum.
// This code snippet sums 1 + 3 + 5 + ... + usernum
// The answer is stored in the variable summedvalue.

N = (int) (usernum+1)/2; // maximum number of integers to be summed
int *v = malloc(N*sizeof(int)); // allocate storage for array v

// Calculate the number of loop counts and assign array v..
count = 0;
k = 1;
while (1) {
if (k>usernum) { // do not extend v beyond usernum
break;
}
v(count) = k; // assign an odd integer to v, including usenum
count++;
k += 2; // k is an odd number
if k>usernum { // handle usernum as odd or even
k = usernum;
}
}
n = count; // the size of array v.

// Calculate the sum in a for loop
summedvalue = 0; // initialize summedvalue
for (i=0; i<=n; i++) {
summedvalue += v(i);
}