The implementation of the function of exponentiation. C / C ++

Maybe this is not the best way, but it works!

long int pow(long int x, unsigned int n) { if (n==0) return 1; else if (n==1) return x; else if (n % 2 == 0 ) return pow( x * x, n/2); else return pow( x * x, n /2)*x; } 

Clarify the question, or the previous answer will be correct. If it is necessary to calculate the real power of a number, then the formula b ^ x = exp (x * ln (b)). If it is necessary to realize both the functions of the exponent and the natural logarithm, please decompose them according to Taylor .

Comment to all the answers. What to do with overflow? Nowhere is it analyzed.

For example library function

 double pow(double x, double y) 

sets errno to erange

Shortest code (:

 class big{/*реализация длинной арифметики*/}; big BinPow(big a, long long n){ big res = big(1); // тут res = 1 while (n) n & 1 ? (res *= a, --n) : (a *= a, n >> 1); reutn res; } 

In general, properties are used here:
1. a ^ n = a ^ (n / 2) * a ^ (n / 2) - for odd n;
2. a ^ n = a ^ (n / 1) * a - for even n.

Long for pathos (:

I would implement the same algorithm as in the first answer, but iteratively, without spending too much time on the recursive call and O (log n) memory in the call stack, and with small optimizations. The following code works only for nonnegative integer n, for other values ​​of n, use Taylor expansion.

 long int pow(long int x, unsigned int n) { long int a = x, p = 1; while (n > 0) { if ((n & 1) != 0) p *= a; a *= a; n >>= 1; } return p; } 

It can be much simpler:

 int pwr (register int m, register int e) { register int temp; temp = 1; for( ; e; e--) tempс= temp * m; return temp; } 

 int pow(int a,int n) { if(n==0) return 1; else if(n==1) return a; else return pow(a+a, n-1); }