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Finding Binary Exponent of Given Number in Golang
• Last Updated : 01 Apr, 2020

Go language provides inbuilt support for basic constants and mathematical functions to perform operations on the numbers with the help of the math package. You are allowed to find the binary exponent of the specified number with the help of Logb() function provided by the math package. So, you need to add a math package in your program with the help of the import keyword to access Logb() function.

Syntax:

`func Logb(a float64) float64`
• If you pass +Inf or -Inf in this function, then this function will return +Inf.
• If you pass 0 in this function, then this function will return -Inf.
• If you pass NaN in this function, then this function will return NaN.

Example 1:

 `// Golang program to illustrate how to find the``// binary exponent of the given number`` ` `package main`` ` `import (``    ``"fmt"``    ``"math"``)`` ` `// Main function``func main() {`` ` `    ``// Finding binary exponent ``    ``// of the given number``    ``// Using Logb() function``    ``res_1 := math.Logb(0)``    ``res_2 := math.Logb(1)``    ``res_3 := math.Logb(math.Inf(1))``    ``res_4 := math.Logb(math.NaN())``    ``res_5 := math.Logb(36)`` ` `    ``// Displaying the result``    ``fmt.Printf(``"Result 1: %.1f"``, res_1)``    ``fmt.Printf(``"\nResult 2: %.1f"``, res_2)``    ``fmt.Printf(``"\nResult 3: %.1f"``, res_3)``    ``fmt.Printf(``"\nResult 4: %.1f"``, res_4)``    ``fmt.Printf(``"\nResult 5: %.1f"``, res_5)`` ` `}`

Output:

```Result 1: -Inf
Result 2: 0.0
Result 3: +Inf
Result 4: NaN
Result 5: 5.0
```

Example 2:

 `// Golang program to illustrate how to find the``// binary exponent of the given number`` ` `package main`` ` `import (``    ``"fmt"``    ``"math"``)`` ` `// Main function``func main() {`` ` `    ``// Finding binary exponent``    ``// of the given number``    ``// Using Logb() function``    ``nvalue_1 := math.Logb(100)``    ``nvalue_2 := math.Logb(26)``    ``res := nvalue_1 + nvalue_2``    ``fmt.Printf(``"%.5f + %.5f = %.5f"``,``            ``nvalue_1, nvalue_2, res)`` ` `}`

Output:

`6.00000 + 4.00000 = 10.00000`

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