CODEWITHSUNDEEP
c++functionloopsfor-loopdefinition
Ximz
Ximz

Reputation: 87

Passing by reference in C++ function

I am writing a simple program to calculate 6^5 using a function. Here's my code

#include <iostream>
using namespace std;

int raiseToPower(int &base, int exponent){
 for (int i = 1; i < exponent; i = i + 1){
   base= base * base;
   }
 return base;
}

int main() {
 int base = 6;
 cout << "6^5 is " << raiseToPower(base, 5)<< endl;

 return 0;
}

The result is

 6^5 is -683606016

instead of 7776

Can you guys explain to me why I got this result? I am not looking for another way to write this code. I know there are simpler ways but I am trying to understand what went wrong

Upvotes: 1

Views: 128

Answers (4)

ObliteratedJillo
ObliteratedJillo

Reputation: 5156

Your calculation is incorrect. The value of variable base should not change. Try this:

 int raiseToPower(int &base, int exponent){
   int result = 1;
   for (int i = 1; i <= exponent; i = i + 1){
      result = result * base;
    }
    return result;
 }

Upvotes: 2

Vlad from Moscow
Vlad from Moscow

Reputation: 310930

Let's consider this loop

 for (int i = 1; i < exponent; i = i + 1){
   base= base * base;
   }

After the first iteration when i is equal to 1 you have

   base= base * base;

that is the result is base ^ 2.

When i = 2 you have

   base= base^2 * base^2;

that is the result is base^4.

When i is equal to 3 you have

   base= base^4 * base^4;

that is the result is base^8.

When i is equal to 4 you have

   base= base^8 * base^8;

that is the result is base^16.

It seems the resulted value is too big to be accomodated in an object of the type int.

Also it is not a good idea when an argument is passed by reference. And the second parameter should have the type unsigned int.

Here is a demonstrative program that shows how the function can be implemented.

#include <iostream>

long long int raiseToPower( int base, unsigned int exponent )
{
    long long int result = 1;

    while ( exponent-- )
    {
        result *= base;
    }

    return result;
}

int main() 
{
    int base = 6;

    std::cout << "6^5 is " << raiseToPower(base, 5) << std::endl;

    return 0;
}

Its output is

6^5 is 7776

Upvotes: 2

Gerhard Stein
Gerhard Stein

Reputation: 1563

Why are you using pass by reference. Setting as much const as possible reduces the odd of making mistakes:

#include <iostream>
using namespace std;

int raiseToPower(const int base, const int exponent) {
    int result = base;
    for (int i = 1; i < exponent; i = i + 1){
        result = result * base;
    }
    return result;
}

int main() {
    int base = 6;
    cout << "6^5 is " << raiseToPower(base, 5)<< endl;

    return 0;
}

If it has to be pass by reference you could do this:

#include <iostream>
using namespace std;

void raiseToPower(const int base,
                 const int exponent,
                 int &result)
{
    result = base;
    for (int i = 1; i < exponent; i = i + 1){
        result = result * base;
    }
}

int main() {
    int base = 6;
    int result;
    raiseToPower(base, 5, result);

    cout << "6^5 is " << result << endl;

    return 0;
}

Upvotes: 3

DeducibleSteak
DeducibleSteak

Reputation: 1498

You probably want your loop like this:

int result = 1;
for (int i = 0; i < exponent; i = i + 1) {
    result = result * base;
}
return result;

Also, you are taking your base parameter to raiseToPower by reference - which means you are changing the base variable in main - probably not what you want to do.

Upvotes: 2

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