Iteration strange output

[fizanimtiaz93]

Hey guys, this code works perfectly fine if n is 0 or 1, except for the fact that when it is any higher, at the end of every output is the square of x. This is the function:

int power1(int x, int n){

    int i = 0;
    if (n == 0){
        return 1;
        i++;}
    if (n == 1){
        return x;
        i++;}
    else {
        unsigned int temp;
        for (temp = x; n > 1; n--){
            temp *= x;
            i++;
            return temp;
        }
        cout << "# of multi's: " << i << endl;
    }
}

And the output is, for example a base number 5 and exponent 14:

5
14
25

[Danny Toledo]

The return statement on line 15 should be after the loop, not in it. You also have a bunch of statements that will never be reached and therefore never executed (lines 6, 9 and 17).

Finally, 5¹⁴ is too big to fit into an integer (so the result will be incorrect). Try some smaller numbers (5⁵, 10⁴, 3⁶ etc...).

[Duthomhas]

Swap lines 15 and 16.

[fizanimtiaz93]

Oh, sorry. I just put in 14 for emphasis. The suggestion of yours works, and I even moved cout << "# of multi's: " << i << endl; to after the brace in line 18, but I still can't get it to print the number of multiplications done in the process. Any suggestions?

[Danny Toledo]

You need to move the cout statement to somewhere before the return statement (but also after the loop).

[Duthomhas]

Remember that when you return from a function, nothing else happens in that function. Program control returns to where the function was called.

So return temp; must come after cout << stuff; if you want to see that stuff.

Hope this helps.

[fizanimtiaz93]

Yeah I changed it to what you guys said, and the output comes out fine, but is there not a way to make the calculated answer come first and then the number of multiplications?

[Danny Toledo]

What is a power? a^b is a multiplied by itself b times. So, 3⁴ = 3 * 3 * 3 * 3. Now, your code would say that the# of multiplications is 3 (in the example, count the number of asterisks/ multiplication operators). So, if that is your desired output, simply print the exponent minus 1:

Note that the first output is the one that you have inside of the power1 function

1 2 3 4 5

b = 3; p = 4; result = power1(b, p); cout<< result<< '\n'; cout<< "mults: "<< (p-1)<< '\n';

# of mult's: 3 81 mults: 3

No need to count the number of multiplications when we can easily calculate it :D

[fizanimtiaz93]

Well that works lol. The program runs just like I'd want it to now. Thanks a bunch for all the help!