golang：对于float64，fmt.Printf中的“％b”有什么作用？在float64中，二进制格式的Min次正规正整数是什么？

Question

golang doc for Package fmt Floating-point and complex constituents says:

Floating-point and complex constituents:
%b decimalless scientific notation with exponent a power of two, in the manner of strconv.FormatFloat with the 'b' format, eg -123456p-78

and my test:

fmt.Printf("%b\n", 1.0) 

the result:
4503599627370496p-52
what is it? your guess is mine!
so test this:

fmt.Printf("0b%b\n", 255) //0b11111111

it is ok. so i think this is interesting and informative to share.

Answer 1

then i did some Research and after many hours of research with IEEE 754 binary representation; many inserting results came up:
good point to start is:

https://en.wikipedia.org/wiki/Double-precision_floating-point_format https://en.wikipedia.org/wiki/IEEE_floating_point

then some test and results:

package main
import (
    "fmt"
    "math"
    "strconv"
       )

func main() {
fmt.Printf("0b%b\n", 255) //0b11111111

fmt.Printf("%b\n", 1.0)               //4503599627370496p-52
fmt.Printf("%#X\n", 4503599627370496) //0X10000000000000
//float64: 1.0 = binary: 0X3FF0000000000000
//so 4503599627370496*2**-52 =("1"+"Fraction")*2**-52
//=0X10 0000 0000 0000*2**-52=  2**52 * 2**-52 = 2**0 = 1=   significand
//2**52=0x10 0000 0000 0000

//1bit=sign 11bit=exponent-biased 52bit)=64 bit:
fmt.Printf("%#X\n", math.Float64bits(1.0)) //1.0=0X3FF0000000000000
//Exp: 0x3FF=1023=E(0) :bias=1023 Emin=1 Emax=2046    Exp=pow(2,0x3FF - 1023)=pow(2,0)=1
//significant: 1.mantisa (53bit nice!) 1.0000000000000

//1.0000000000000002, the smallest number > 1
fmt.Printf("%#X\n", math.Float64bits(1.0000000000000002)) //0X3FF0000000000001

// 1.0000000000000004, the next numer after 1.0000000000000002
fmt.Printf("%#X\n", math.Float64bits(1.0000000000000004)) //0X3FF0000000000002

fmt.Printf("%#X\n", math.Float64bits(2.0))  //0X4000000000000000
fmt.Printf("%#X\n", math.Float64bits(-2.0)) //0XC000000000000000

// Min subnormal positive double
fmt.Printf("%v\n", math.Float64frombits(1))   //5e-324
fmt.Printf("%#X\n", math.Float64bits(5e-324)) //0X0000000000000001
//Exp(2,-1022-52)=Exp(2,-1074)=5e-324

//Max subnormal double
fmt.Printf("%v\n", math.Float64frombits(0x000fffffffffffff)) //2.225073858507201e-308

fmt.Printf("%v\n", math.Float64frombits(0X0000000000000000)) //0
fmt.Printf("%v\n", math.Float64frombits(0X8000000000000000)) //-0
fmt.Printf("%v\n", math.Float64frombits(0X7FF0000000000000)) //+Inf
fmt.Printf("%v\n", math.Float64frombits(0XFFF0000000000000)) //-Inf
fmt.Printf("%v\n", math.Float64frombits(0x7fffffffffffffff)) //NaN

fmt.Printf("%#X\n%[1]b\n", math.Float64bits(0.1)) //0X3FB 999999999999A
//0 1111111011 1001100110011001100110011001100110011001100110011010

fmt.Printf("%#X\n%[1]b\n", math.Float64bits(0.2)) //0X3FC999999999999A
//11111111001001100110011001100110011001100110011001100110011010

fmt.Printf("%#X\n%[1]b\n", math.Float64bits(0.3)) //0X3FD3333333333333
//11111111010011001100110011001100110011001100110011001100110011
fmt.Println(1.0 / 3.0) //0.3333333333333333
//By default, 1/3 rounds down, instead of up like single precision,
//because of the odd number of bits in the significand
fmt.Printf("%#X\n%[1]b\n", math.Float64bits(1.0/3.0)) //0X3FD5555555555555
//11111111010101010101010101010101010101010101010101010101010101
/*
   Given the hexadecimal representation 3FD5 5555 5555 555516,
     Sign = 0
     Exponent = 0x3FD = 1021
     Exponent Bias = 1023 (constant value)
     Fraction = 5 5555 5555 555516
     Value = 2(Exponent − Exponent Bias) × 1.Fraction // Note that Fraction must not be converted to decimal here
           = 2**−2 × (15 5555 5555 555516 × 2**−52)
           = 2**−54 × 15 5555 5555 555516
           = 0.333333333333333314829616256247390992939472198486328125
           ≈ 1/3
*/

var f float64 = 0.1
var bits uint64 = math.Float64bits(f) //IEEE 754 binary representation of f
fmt.Printf("%#X %[1]b\n", bits)
//0X3FB999999999999A 11111110111001100110011001100110011001100110011001100110011010

fmt.Printf("%b\n", f) //7205759403792794p-56
fmt.Printf("%b  %b\n", 7205759403792794, -56)
//11001100110011001100110011001100110011001100110011010  111000
fmt.Println(len("11001100110011001100110011001100110011001100110011010"))
//text search in this text=> 53 bit right side

// 1 11111110101 100011110110001111100111100101011000111010000110011
fmt.Printf("-: %b\n", math.Float64bits(-0.1e+308)) // so left bit is sign bit
//1 111111110101 100011110110001111100111100101011000111010000110011
fmt.Printf("exp: %b\n", math.Float64bits(+0.1e-308))
//1 01110000001 010101110010011010001111110110101111
// 11Exponent bits

//  12345678901
i, err := strconv.ParseInt("11111110101", 2, 64) //2037
fmt.Println("E", i-1023)                         //1014

fmt.Printf("%b\n", 0.2) //7205759403792794p-55
fmt.Printf("%b\n", 0.3) //5404319552844595p-54

n, err := fmt.Printf("%b %b\n", 1.0, math.Float64bits(1.0))
//4503599627370496p-52 11111111110000000000000000000000000000000000000000000000000000
fmt.Println(n, err) //84 <nil>
//no err
fmt.Printf("'%[1]*.[2]*[3]f'\n", 12, 4, 1234.1234) //'   1234.1234'

}

conclusion:
%b for float shows only significand , done.
this single line took many hours of my time!
any comments are welcome, thanks.