c++ - Implement pow(x, n) -


why function give wrong answer -1 rather right answer 1 when try this? mypow(-1.00000, -2147483648)

double quickpower(double x, int n)  {     if(n==0){         return 1;     }      if(n==1){         return x;     }      if(n>=2){         int res=n%2;         double half=quickpower(x,n/2);         return res? half*half*x: half*half;     } }  double mypow(double x, int n) {     return n>=0? quickpower(x,n):(1/quickpower(x,-n)); }

i try run code below. "hello world" printed out. here did't specify data type still pass if statement. why?
if (-1 > 2147483648) { cout << "hello world"; }

your problem due integer overflow when calculating -n. on system (and local one) int_min=-2147483648 , int_max=2147483647.

so problem -(int_min) not representable integer. can avoid issue without going higher precision integer type:

since

xn = xn+1 / x = (1/x) / x-(n+1)

we can rewrite mypow as

double mypow(double x, int n) {     return n>=0? quickpower(x,n):(1/x)/quickpower(x,-(n+1)); }

this function ok since -(int_min+1)=int_max.

it's worth noting have mypow(0,-k) return either +/- infinity (n=-1) or nan (n<-1). if need case consistent little more work required. in general handling of infinite / nan values tricky pow (and not "correct" in this, or original implementation) - worth man page c pow function edge cases.


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