Reputation: 149
Calculate nth power of P (both p and n are positive integer) using a recursive function myPowerFunction(int p, int n, int ¤tCallNumber). currentCallNumber is a reference parameter and stores the number of function calls made so far. myPowerFunction returns the nth power of p.
int myPowerFunction(int p, int n, int &z)
{
z++;
if(n==1)return p;
else if(n==0)return 1;
else if(n%2==0)return myPowerFunction(p,n/2,z)*myPowerFunction(p,n/2,z);
else return myPowerFunction(p,n/2,z)*myPowerFunction(p,n/2,z)*p;
}
int main()
{
cout << myPowerFunction(3,4,1);
}
Upvotes: 14
Views: 33994
Reputation: 491
You don't have to give a reference as parameter as many here state.
But yes, your input for z
cannot be modified as it comes from read-only memory. Treat the input for z
as const
, copy z
internally and give the copy as reference. Then your desired usage works:
int myPowerFunction(int p, int n, const int &z) // z is now const !
{
int _z = z + 1; // copy !
if (n == 1) return p;
else if (n == 0) return 1;
else if (n % 2 == 0) return myPowerFunction(p, n /2 , _z) * myPowerFunction(p, n / 2, _z);
else return myPowerFunction(p, n / 2, _z) * myPowerFunction(p, n / 2, _z) * p;
}
int main()
{
std::cout << myPowerFunction(3, 4, 1);
}
Upvotes: 4
Reputation: 180500
In myPowerFunction(3,4,1)
the literal 1
cannot be passed to a non const
reference as it is a prvalue [basic.lval]. You need to store the value into a variable and then use that variable when calling the function.
int z = 0;
std::cout << myPowerFunction(3, 4, z);
Upvotes: 2
Reputation: 44238
Third parameter expects a lvalue, so you cannot pass numeric constant there, so possible solution can be:
int z = 1;
cout<< myPowerFunction(3,4,z);
or better to create a function that calls recursive one:
int myPowerFunction(int p, int n)
{
int z = 1;
return myPowerFunction(p,n,z);
}
Upvotes: 2
Reputation: 76245
You need a variable to pass as the third argument in main_program. You can't pass a constant as a non-const reference.
int count = 0;
std::cout << myPowerFunction(3, 4, count) << 'n';
std::cout << count << '\n';
Upvotes: 15