power of an integer in c++ [duplicate]

Question

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The most efficient way to implement an integer based power function pow(int, int) (17 answers)

Closed 4 years ago.

I need to get the result from pow(a,b) as an integer (both a and b are integers too). currently the calculations where (int) pow( (double)a, (double)b) is included are wrong. Maybe someone can help with a function that does the pow(a,b) with integers and returns an integer too?

But here is the odd part: I made my script in Linux with Geany (and g++/gcc compiler) and had just pow(a,b) the script compiled and worked fine. But in university I have Dev-C++ (and MS Windows). In Dev-C++ the script didn't compile with an error [Warning] converting toint' from double'

I need to make this scrpit work under Windows (and Mingw compiler) too.

Answer 1

There are two alternatives here, when we want to count power(a,n) we may write code which is very short and works in O(logn) time, but is recursively and therefore requires creating new stackframe for each call and needs a bit more time than loop iteration. So short code is:

int power(int a, int n){
    if(n == 0) return 1;
    int keep = power(a,n/2);
    if(n & 1) return keep*keep*a;   // (n & 1) is and operation of 1 and the 
    return keep*keep;               // last bit of n
}

and as for the faster code, here it is using while loop:

int power(int a, int n) {
    int res = 1;
    while (n) {
        if (n & 1)
            res *= a;
        a *= a;
        n >>= 1;
    }
    return res;
}

Answer 2

A nice recursive approach you can show off:

int myPow(int x, int p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  return x * myPow(x, p-1);
}

Answer 3

C++ standard doesn't have int pow(int, int) (It has double pow(double, int), float ...). Microsoft's cmath uses C math.h that has not ipow. Some cmath headers define template version of pow.

$ cat main.cpp
#include <cmath>

int main() {
  std::pow(2,2);
}

$ gcc main.cpp # this cmath has template pow
...snip... std::pow<int, int>(int, int)]+0x16): undefined reference to `pow'
collect2: ld returned 1 exit status
1 ;( user@host:
$ gcc main.cpp -lm

Search for function:ipow lang:c++ on Google Code .

Here's example from the first link:

template <typename Type1, typename Type2>
Type1 ipow(Type1 a, Type2 ex)
// Return a**ex
{
    if ( 0==ex )  return 1;
    else
    {
        Type1 z = a;
        Type1 y = 1;
        while ( 1 )
        {
            if ( ex & 1 )  y *= z;
            ex /= 2;
            if ( 0==ex )  break;
            z *= z;
        }
        return y;
    }
}

See calculating integer powers (squares, cubes, etc.) in C++ code.

Answer 4

Binary powering, aka exponentiation by squaring.

int powi (int base, unsigned int exp)
{
    int res = 1;
    while (exp) {
        if (exp & 1)
            res *= base;
        exp >>= 1;
        base *= base;
    }
    return res;
}

Note that this returns 1 for powi(0,0).

Answer 5

A better recursive approach than Zed's.

int myPow(int x, int p)
{
  if (p == 0) return 1;
  if (p == 1) return x;

  int tmp = myPow(x, p/2);
  if (p%2 == 0) return tmp * tmp;
  else return x * tmp * tmp;
}

Much better complexity there O(log²(p)) instead of O(p).

Answer 6

I assume your homework assignment is to write an integral exponent function. First, take a look at what an exponent is:

http://en.wikipedia.org/wiki/Exponent

Then, look in your textbook for how to multiply numbers in C. You'll want to use a for loop.