Converting from an integer to its binary representation

Question

Has anyone got an idea if there is any inbuilt functionality in Go for converting from any one of the numeric types to its binary number form.

For example, if

Answer 1

The strconv package has FormatInt, which accepts an int64 and lets you specify the base.

n := int64(123)

fmt.Println(strconv.FormatInt(n, 2)) // 1111011

DEMO: http://play.golang.org/p/leGVAELMhv

http://golang.org/pkg/strconv/#FormatInt

func FormatInt(i int64, base int) string

FormatInt returns the string representation of i in the given base, for 2 <= base <= 36. The result uses the lower-case letters 'a' to 'z' for digit values >= 10.

Answer 2

This code works on big integers *big.Int :

x := big.NewInt(123)
s := fmt.Sprintf("%b", x)
// s == "1111011"

because *big.Int implements the fmt.Formatter interface.

Taken from https://stackoverflow.com/a/23317788/871134

Answer 3

Unsafe pointers must be used to correctly represent negative numbers in binary format.

package main

import (
    "fmt"
    "strconv"
    "unsafe"
)

func bInt(n int64) string {
    return strconv.FormatUint(*(*uint64)(unsafe.Pointer(&n)), 2)
}

func main() {
    fmt.Println(bInt(-1))
}

https://play.golang.org/p/GxXjjWMyC4x

Answer 4

package main

import . "fmt"

func main(){
    Printf("%d == %08b\n",0,0)
    Printf("%d == %08b\n",1,1)
    Printf("%d == %08b\n",2,2)
    Printf("%d == %08b\n",3,3)
    Printf("%d == %08b\n",4,4)
    Printf("%d == %08b\n",5,5)
}

results in:

0 == 00000000
1 == 00000001
2 == 00000010
3 == 00000011
4 == 00000100
5 == 00000101

Answer 5

Building on the answer provided by @Mark

Although the OP asked how to print an integer, I often want to look at more then 64 bits worth of data, without my eyes boggling:

/* --- Credit to Dave C in the comments --- */
package main

import (
    "bytes"
    "fmt"
)

func main() {
    fmt.Printf("<%s>\n", fmtBits([]byte{0xDE, 0xAD, 0xBE, 0xEF, 0xF0, 0x0D, 0xDE, 0xAD, 0xBE, 0xEF, 0xF0, 0x0D}))

    // OUTPUT:
    // <11011110 10101101 10111110 11101111 11110000 00001101 11011110 10101101 10111110 11101111 11110000 00001101>
}

func fmtBits(data []byte) []byte {
    var buf bytes.Buffer
    for _, b := range data {
        fmt.Fprintf(&buf, "%08b ", b)
    }
    buf.Truncate(buf.Len() - 1) // To remove extra space
    return buf.Bytes()
}

see this code in play.golang.org

Answer 6

An alternate way for the accepted answer would be to simply do

s := fmt.Sprintf("%b", 123)
fmt.Println(s)                 // 1111011

For a more rich representation you can use the unsafe package(strongly discouraged) as

a := int64(123)
byteSliceRev := *(*[8]byte)(unsafe.Pointer(&a))
byteSlice := make([]byte, 8)
for i := 0; i < 8; i++ {
    byteSlice[i] = byteSliceRev[7 - i]
}
fmt.Printf("%b\n", byteSlice)

This works for negative integers too.