How to calculate Base64 encoded string length?

Question

I've read following topic and found this formula there:

length = 4*(n/3)

I started to test it:

1 symbol: Base64.getEncoder().encodeToString("1".getBytes()) =>MQ==(4 symbols)

2 symbols: Base64.getEncoder().encodeToString("12".getBytes()) =>MTI= (4 symbols)

5 symbols: Base64.getEncoder().encodeToString("12345".getBytes()) =>MTIzNDU=(8 symbols)

8 symbols: Base64.getEncoder().encodeToString("12345678".getBytes()) =>MTIzNDU2Nzg=(12 symbols)

21symbols: Base64.getEncoder().encodeToString("123456789012345678901".getBytes()) => MTIzNDU2Nzg5MDEyMzQ1Njc4OTAx (28 symbols)

Looks like this formula doesn't work.

Can you please explain mu results?

It looks like you should take to ceil of (n/3). e.g - Ceil(8/3) = 3. To for your forth example the calculation will be- 4 * Ceil(8/3) = 4 * 3 = 12 — Kram, Apr 03 '18 at 15:05
With ceil, it's a *bound* rather than an exact result. It's a guide to aid scaling calculations. — Bohemian, Apr 03 '18 at 15:06
22 symbols: **Base64.getEncoder().encodeToString("1234567890123456789012".getBytes()) =>*MTIzNDU2Nzg5MDEyMzQ1Njc4OTAxMg==(32 symbols) — gstackoverflow, Apr 03 '18 at 15:10
@Bohemian Do you happen to have an example when `4*ceil(n/3)` does not give an exact result? — lexicore, Apr 03 '18 at 15:18

Joop Eggen · Answer 1 · 2018-04-03T15:41:35.607

4

With 64 digits (2⁶) one digit can represent 6 bits. Hence 4 digits can represent exactly 4*6 bits = 3 bytes.

(Using ÷ for explicit integer division:)

With n bytes, 4*(n÷3) digits are needed plus for the remainder n%3 (0 < 3 bytes) there is a need for 0 upto 4 digits:

0 bytes (0 bits)    0 digits
1 byte  (8 bits)    2 digits (12 bits)     + "=="
2 bytes (16 bits)   3 digits (18 bits)     + "="

Often there is a padding upto 4 digits/padding chars, using =. This cannot be 0 as one then would add a byte 0x0.

Then the formula is 4 * Math.ceil(n / 3.0).

Without padding: Math.ceil(n * 8 / 6.0) = Math.ceil(n * 4 / 3.0) = (n * 4 + (3 - 1)) ÷ 3.

In java one should use int division only:

int base64Length(byte[] b) {
    int n = b.length;
    return (n * 4 + 2)/3;
}

edited Apr 03 '18 at 15:41

answered Apr 03 '18 at 15:29

Joop Eggen

I understood 4 * Math.ceil(n / 3.0). Bit it is not cleat where did take (12 bits) and (18 bits) – gstackoverflow Apr 03 '18 at 15:41
@gstackoverflow for 1 byte one needs 2 digits (8 <= 12 bit) and for 2 bytes one needs 3 digits (16 bit <= 18 bit). – Joop Eggen Apr 03 '18 at 15:43
4*Math.ceil(16 / 3.0) == 20 – gstackoverflow Apr 03 '18 at 15:47
`4*Math.ceil(16 / 3.0) = 4 * ceil(5.333) = 4 * 6 = 24` Ceil being round-up. – Joop Eggen Apr 03 '18 at 15:54
you are correct but how did you calculate this line: **2 bytes (16 bits) 3 digits (18 bits) + "=" **? – gstackoverflow Apr 03 '18 at 16:02
To represent one byte (8 bits) with digits, one needs at 2 digits à 6 bits = 12 bits. One digit could only represent at most 6 bits payload. From those 12 bits, 4 bits are free: in effect one of 16 different digits would do. But in general the free bits are zeroed. – Joop Eggen Apr 03 '18 at 16:25

Steve11235 · Answer 2 · 2018-04-04T16:08:19.917

-2

Try length = 4*((n/3) + 1), where "/" is integer division.

Edit: Lexicore is correct, my formula fails when the remainder is zero.

int length = 4 * (n / 3);
if (n % 3 > 0) length++;

edited Apr 04 '18 at 16:08

answered Apr 03 '18 at 15:28

Steve11235

`4*((n/3) + 1)` gives `32` for `n=21` whereas the correct answer is `28`. – lexicore Apr 03 '18 at 15:37

2 Answers2