Why does Go use bit operation in base64 encode?

Code 2 is base64.Encode of Go.
This code uses bit operation many times.

For example,

val := uint(src[si+0])<<16 | uint(src[si+1])<<8 | uint(src[si+2])

And,

val := uint(src[si+0])>>18&0x3F]

I don't know why these bit operations are need to encode to base64.
What is the meaning of these bit operations?

Code:

func (enc *Encoding) Encode(dst, src []byte) {
if len(src) == 0 {
return
}
di, si := 0, 0
n := (len(src) / 3) * 3
for si < n {
// Convert 3x 8bit source bytes into 4 bytes
val := uint(src[si+0])<<16 | uint(src[si+1])<<8 | uint(src[si+2])
dst[di+0] = enc.encode[val>>18&0x3F]
dst[di+1] = enc.encode[val>>12&0x3F]
dst[di+2] = enc.encode[val>>6&0x3F]
dst[di+3] = enc.encode[val&0x3F]
si += 3
di += 4
}
remain := len(src) - si
if remain == 0 {
return
}
// Add the remaining small block
val := uint(src[si+0]) << 16
if remain == 2 {
val |= uint(src[si+1]) << 8
}
dst[di+0] = enc.encode[val>>18&0x3F]
dst[di+1] = enc.encode[val>>12&0x3F]
switch remain {
case 2:
dst[di+2] = enc.encode[val>>6&0x3F]
if enc.padChar != NoPadding {
dst[di+3] = byte(enc.padChar)
}
case 1:
if enc.padChar != NoPadding {
dst[di+2] = byte(enc.padChar)
dst[di+3] = byte(enc.padChar)
}
}
}

Here is a commented Javascript implementation of the same algorithm: https://en.wikibooks.org/wiki/Algorithm_Implementation/Miscellaneous/Base64#Javascript

It tells you that this loop:

• takes three ASCII chars (8-bit): src[si+0], src[si+1] and src[si+2]

• merges them into one 24-bit number (that's val := uint(src[si+0])<<16 | uint(src[si+1])<<8 | uint(src[si+2]))

• re-separate this number into four indices (6-bit) for the base64 character list. (val>>18&0x3F takes the 18th to 24th bit of the previously calculated number, etc.)

You can also read this: https://en.wikipedia.org/wiki/Base64#Examples