在 C/C++ 中快速得到 10 的 n(10^n) 次方

Question

well thanks for clicking in. i want to calculate 10(yes only 10) to the power n[0..308] fast. i came up with some methods.

1)

double f(int n) {
  return pow(10.0, n);
}
double f1(int n) {
  double a = 10.0;
  double res = 1.0;
  while(n) {
    if(n&1) res *= a;
    a *= a;
    n >>= 1;
  }
  return res;
}

time: O(logn), could it be faster? ( // f1() can do a little bit optimization but still O(logn))

2)

double f2(int n) {
  static const double e[] = { 1e+0, 1e+1, 1e+2, ..., 1e+308 };
  return e[n];
}

time: O(1), very good. But space: 309 * 8 bytes = 2472 bytes.. whoops it's too huge...

3)

double f3(int n){
    static const double e[] = {
        1e+1, 1e+2, 1e+4, 1e+8, 1e+16, 1e+32, 1e+64, 1e+128, 1e+256
    };
    double res = 1.0;
    for(int i = 0; n; ++i){
        if(n & 1){
            res *= e[i];
        }
        n >>= 1;
    }
    return res;
}

f3 combines f1 and f2 to avoid multiplication such as 1e128*1e128, i wished it faster, but.. actually f3 is slower than f2.. because of ++ii guess..

well i almost gave up before i typed these codes,

int main(){
    double d = 1e+2;
    return 0;
}

and compiled it to .s by g++

LCPI0_0:
    .quad   0x4059000000000000              ## double 100

HOW does the compiler know 1e+2 is 0x4059000000000000?

i mean all i want to get is a double values 1e+n. but when a compiler compiles "double d = 1e+2", it know d should be 0x4059000000000000. can i just use some method to directly return something like 1e+n. OR can i do some things beyond C/C++ to get my value?

thanks a lot. Please point out if there is something wrong or unclear.

Answer 1

Have you tried std::pow(std::valarray) ? 🤔