'How to use Math.Pow with integers in Golang
I keep getting the error "cannot use a (type int) as type float64 in argument to math.Pow, cannot use x (type int) as type float64 in argument to math.Pow, invalid operation: math.Pow(a, x) % n (mismatched types float64 and int)"
func pPrime(n int) bool {
var nm1 int = n - 1
var x int = nm1/2
a := 1;
for a < n {
if (math.Pow(a, x)) % n == nm1 {
return true
}
}
return false
}
Solution 1:[1]
func powInt(x, y int) int {
return int(math.Pow(float64(x), float64(y)))
}
In case you have to reuse it and keep it a little more clean.
Solution 2:[2]
If your inputs are int and the output is always expected to be int, then you're dealing with 32-bit numbers. It's more efficient to write your own function to handle this multiplication rather than using math.Pow. math.Pow, as mentioned in the other answers, expects 64-bit values.
Here's a Benchmark comparison for 15^15 (which approaches the upper limits for 32-bit representation):
// IntPow calculates n to the mth power. Since the result is an int, it is assumed that m is a positive power
func IntPow(n, m int) int {
if m == 0 {
return 1
}
result := n
for i := 2; i <= m; i++ {
result *= n
}
return result
}
// MathPow calculates n to the mth power with the math.Pow() function
func MathPow(n, m int) int {
return int(math.Pow(float64(n), float64(m)))
}
The result:
go test -cpu=1 -bench=.
goos: darwin
goarch: amd64
pkg: pow
BenchmarkIntPow15 195415786 6.06 ns/op
BenchmarkMathPow15 40776524 27.8 ns/op
I believe the best solution is that you should write your own function similar to IntPow(m, n int) shown above. My benchmarks show that it runs more than 4x faster on a single CPU core compared to using math.Pow.
Solution 3:[3]
Since nobody mentioned an efficient way (logarithmic) to do Pow(x, n) for integers x and n is as follows if you want to implement it yourself:
// Assumption: n >= 0
func PowInts(x, n int) int {
if n == 0 { return 1 }
if n == 1 { return x }
y := PowInts(x, n/2)
if n % 2 == 0 { return y*y }
return x*y*y
}
Solution 4:[4]
If you want the exact exponentiation of integers, use (*big.Int).Exp. You're likely to overflow int64 pretty quickly with powers larger than two.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | nikoksr |
| Solution 2 | |
| Solution 3 | Eissa N. |
| Solution 4 | ouflak |
