Converting from an integer to its binary representation

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 123 was the input, the string "1111011" would be the output.


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


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.

See also the fmt package:

n := int64(123)
fmt.Printf("%b", n)  // 1111011

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

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 (

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

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

package main

import (

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

func main() {

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