How the SHA-2 (SHA-256) Hashing Algorithm Works

SHA-2 (Secure Hash Algorithm 2) is one of the most popular families of hashing algorithms. In this article, we’ll walk through each step of the SHA-256 algorithm, which belongs to SHA-2, and show how it works with a real example.

What is a hash function?

If you want to know more about hash functions, you can read Wikipedia. But to understand what this is about, let’s remember the three main purposes of a hash function:

  • ensure data integrity (invariance) check;
  • accept input of any length and output the result of a fixed length;
  • change data irreversibly (input cannot be obtained from output).

SHA-2 and SHA-256

SHA-2 is a family of algorithms with the general idea of ​​hashing data. SHA-256 sets additional constants that define the behavior of the SHA-2 algorithm. One of these constants is the output size. “256” and “512” refer to the respective sizes of the output data in bits.

We’ll look at an example of how SHA-256 works.

SHA-256 “hello world”

Step 1. Pre-processing

1. Let’s convert “hello world” to binary:

01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111
01110010 01101100 01100100

2. Add one unit:

01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111
01110010 01101100 01100100 1

3. Fill in with zeros until the data becomes a multiple of 512 without the last 64 bits (in our case 448 bits):

01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111
01110010 01101100 01100100 10000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000

4. Add 64 bits to the end, where 64 bits is a big-endian integer denoting the length of the input data in binary. In our case 88, in binary — “1011000”.

01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111
01110010 01101100 01100100 10000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
00000000 00000000 00000000 00000000 00000000 00000000 00000000 01011000

Now we have an input that will always be divisible by 512 without remainder.

️️Step 2. Initializing hash values ​​(h)

Let’s create 8 hash values. These are constants representing the first 32 bits of the fractional parts of the square roots of the first 8 primes: 2, 3, 5, 7, 11, 13, 17, 19.

h0 := 0x6a09e667
h1 := 0xbb67ae85
h2 := 0x3c6ef372
h3 := 0xa54ff53a
h4 := 0x510e527f
h5 := 0x9b05688c
h6 := 0x1f83d9ab
h7 := 0x5be0cd19

Step 3. Initialization of rounded constants (k)

Let’s create some more constants, this time there are 64 of them. Each value is the first 32 bits of the fractional parts of the cube roots of the first 64 primes (2–311).

0x428a2f98 0x71374491 0xb5c0fbcf 0xe9b5dba5 0x3956c25b 0x59f111f1 0x923f82a4 0xab1c5ed5
0xd807aa98 0x12835b01 0x243185be 0x550c7dc3 0x72be5d74 0x80deb1fe 0x9bdc06a7 0xc19bf174
0xe49b69c1 0xefbe4786 0x0fc19dc6 0x240ca1cc 0x2de92c6f 0x4a7484aa 0x5cb0a9dc 0x76f988da
0x983e5152 0xa831c66d 0xb00327c8 0xbf597fc7 0xc6e00bf3 0xd5a79147 0x06ca6351 0x14292967
0x27b70a85 0x2e1b2138 0x4d2c6dfc 0x53380d13 0x650a7354 0x766a0abb 0x81c2c92e 0x92722c85
0xa2bfe8a1 0xa81a664b 0xc24b8b70 0xc76c51a3 0xd192e819 0xd6990624 0xf40e3585 0x106aa070
0x19a4c116 0x1e376c08 0x2748774c 0x34b0bcb5 0x391c0cb3 0x4ed8aa4a 0x5b9cca4f 0x682e6ff3
0x748f82ee 0x78a5636f 0x84c87814 0x8cc70208 0x90befffa 0xa4506ceb 0xbef9a3f7 0xc67178f2

Step 4. Main loop

The following steps will be performed for each 512-bit “chunk” of input data. Our test phrase “hello world” is pretty short, so there is only one “chunk”. At each iteration of the loop, we will change the values ​​of the hash functions h0- h7to get the final result.

Step 5. Create a message queue (w)

1. Copy the input from step 1 into a new array, where each record is a 32-bit word:

01101000011001010110110001101100 01101111001000000111011101101111
01110010011011000110010010000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000001011000

2. Add 48 more words, initialized to zero, to get an array w[0…63]:

01101000011001010110110001101100 01101111001000000111011101101111
01110010011011000110010010000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000001011000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
...
...
00000000000000000000000000000000 00000000000000000000000000000000

3. Change the zero indices at the end of the array using the following algorithm:

  • For ifrom w[16…63]:
  • s0 = (w[i-15] rightrotate 7) xor (w[i-15] rightrotate 18) xor (w[i-15] righthift 3)
  • s1 = (w[i-2] rightrotate 17) xor (w[i-2] rightrotate 19) xor (w[i-2] righthift 10)
  • w [i] = w[i-16] + s0 + w[i-7] + s1

Let’s see how this works for w[16]:

w[1] rightrotate 7:
01101111001000000111011101101111 -> 11011110110111100100000011101110
w[1] rightrotate 18:
01101111001000000111011101101111 -> 00011101110110111101101111001000
w[1] rightshift 3:
01101111001000000111011101101111 -> 00001101111001000000111011101101
s0 = 11011110110111100100000011101110 XOR 00011101110110111101101111001000 XOR 00001101111001000000111011101101s0 = 11001110111000011001010111001011w[14] rightrotate 17:
00000000000000000000000000000000 -> 00000000000000000000000000000000
w[14] rightrotate19:
00000000000000000000000000000000 -> 00000000000000000000000000000000
w[14] rightshift 10:
00000000000000000000000000000000 -> 00000000000000000000000000000000
s1 = 00000000000000000000000000000000 XOR 00000000000000000000000000000000 XOR 00000000000000000000000000000000s1 = 00000000000000000000000000000000w[16] = w[0] + s0 + w[9] + s1w[16] = 01101000011001010110110001101100 + 11001110111000011001010111001011 + 00000000000000000000000000000000 + 00000000000000000000000000000000// сложение рассчитывается по модулю 2^32w[16] = 00110111010001110000001000110111

This leaves us 64 words in our message queue ( w):

01101000011001010110110001101100 01101111001000000111011101101111
01110010011011000110010010000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000000000000
00000000000000000000000000000000 00000000000000000000000001011000
00110111010001110000001000110111 10000110110100001100000000110001
11010011101111010001000100001011 01111000001111110100011110000010
00101010100100000111110011101101 01001011001011110111110011001001
00110001111000011001010001011101 10001001001101100100100101100100
01111111011110100000011011011010 11000001011110011010100100111010
10111011111010001111011001010101 00001100000110101110001111100110
10110000111111100000110101111101 01011111011011100101010110010011
00000000100010011001101101010010 00000111111100011100101010010100
00111011010111111110010111010110 01101000011001010110001011100110
11001000010011100000101010011110 00000110101011111001101100100101
10010010111011110110010011010111 01100011111110010101111001011010
11100011000101100110011111010111 10000100001110111101111000010110
11101110111011001010100001011011 10100000010011111111001000100001
11111001000110001010110110111000 00010100101010001001001000011001
00010000100001000101001100011101 01100000100100111110000011001101
10000011000000110101111111101001 11010101101011100111100100111000
00111001001111110000010110101101 11111011010010110001101111101111
11101011011101011111111100101001 01101010001101101001010100110100
00100010111111001001110011011000 10101001011101000000110100101011
01100000110011110011100010000101 11000100101011001001100000111010
00010001010000101111110110101101 10110000101100000001110111011001
10011000111100001100001101101111 01110010000101111011100000011110
10100010110101000110011110011010 00000001000011111001100101111011
11111100000101110100111100001010 11000010110000101110101100010110

Step 6. Compression cycle

  1. We initialize the variables a, b, c, d, e, f, g, hand set them equal to the current hash values, respectively. h0, h1, h2, h3, h4, h5, h6, h7...
  2. Let’s start a compression cycle that will change the values ​​of a… h. The loop looks like this:
  • for i from 0 to 63
  • S1 = (e rightrotate 6) xor (e rightrotate 11) xor (e rightrotate 25)
  • ch = (e and f) xor ((not e) and g)
  • temp1 = h + S1 + ch + k[i] + w[i]
  • S0 = (a rightrotate 2) xor (a rightrotate 13) xor (a rightrotate 22)
  • maj = (a and b) xor (a and c) xor (b and c)
  • temp2 := S0 + maj
  • h = g
  • g = f
  • e = d + temp1
  • d = c
  • c = b
  • b = a
  • a = temp1 + temp2

Let’s go through the first iteration. The addition is calculated modulo 2 ^ 32:

a = 0x6a09e667 = 01101010000010011110011001100111
b = 0xbb67ae85 = 10111011011001111010111010000101
c = 0x3c6ef372 = 00111100011011101111001101110010
d = 0xa54ff53a = 10100101010011111111010100111010
e = 0x510e527f = 01010001000011100101001001111111
f = 0x9b05688c = 10011011000001010110100010001100
g = 0x1f83d9ab = 00011111100000111101100110101011
h = 0x5be0cd19 = 01011011111000001100110100011001
e rightrotate 6:
01010001000011100101001001111111 -> 11111101010001000011100101001001
e rightrotate 11:
01010001000011100101001001111111 -> 01001111111010100010000111001010
e rightrotate 25:
01010001000011100101001001111111 -> 10000111001010010011111110101000
S1 = 11111101010001000011100101001001 XOR 01001111111010100010000111001010 XOR 10000111001010010011111110101000
S1 = 00110101100001110010011100101011
e and f:
01010001000011100101001001111111
& 10011011000001010110100010001100 =
00010001000001000100000000001100
not e:
01010001000011100101001001111111 -> 10101110111100011010110110000000
(not e) and g:
10101110111100011010110110000000
& 00011111100000111101100110101011 =
00001110100000011000100110000000
ch = (e and f) xor ((not e) and g)
= 00010001000001000100000000001100 xor 00001110100000011000100110000000
= 00011111100001011100100110001100
// k[i] is the round constant
// w[i] is the batch
temp1 = h + S1 + ch + k[i] + w[i]
temp1 = 01011011111000001100110100011001 + 00110101100001110010011100101011 + 00011111100001011100100110001100 + 1000010100010100010111110011000 + 01101000011001010110110001101100
temp1 = 01011011110111010101100111010100
a rightrotate 2:
01101010000010011110011001100111 -> 11011010100000100111100110011001
a rightrotate 13:
01101010000010011110011001100111 -> 00110011001110110101000001001111
a rightrotate 22:
01101010000010011110011001100111 -> 00100111100110011001110110101000
S0 = 11011010100000100111100110011001 XOR 00110011001110110101000001001111 XOR 00100111100110011001110110101000
S0 = 11001110001000001011010001111110
a and b:
01101010000010011110011001100111
& 10111011011001111010111010000101 =
00101010000000011010011000000101
a and c:
01101010000010011110011001100111
& 00111100011011101111001101110010 =
00101000000010001110001001100010
b and c:
10111011011001111010111010000101
& 00111100011011101111001101110010 =
00111000011001101010001000000000
maj = (a and b) xor (a and c) xor (b and c)
= 00101010000000011010011000000101 xor 00101000000010001110001001100010 xor 00111000011001101010001000000000
= 00111010011011111110011001100111
temp2 = S0 + maj
= 11001110001000001011010001111110 + 00111010011011111110011001100111
= 00001000100100001001101011100101
h = 00011111100000111101100110101011
g = 10011011000001010110100010001100
f = 01010001000011100101001001111111
e = 10100101010011111111010100111010 + 01011011110111010101100111010100
= 00000001001011010100111100001110
d = 00111100011011101111001101110010
c = 10111011011001111010111010000101
b = 01101010000010011110011001100111
a = 01011011110111010101100111010100 + 00001000100100001001101011100101
= 01100100011011011111010010111001

All calculations are performed 63 more times, changing the variables аh. As a result, we should get the following:

h0 = 6A09E667 = 01101010000010011110011001100111
h1 = BB67AE85 = 10111011011001111010111010000101
h2 = 3C6EF372 = 00111100011011101111001101110010
h3 = A54FF53A = 10100101010011111111010100111010
h4 = 510E527F = 01010001000011100101001001111111
h5 = 9B05688C = 10011011000001010110100010001100
h6 = 1F83D9AB = 00011111100000111101100110101011
h7 = 5BE0CD19 = 01011011111000001100110100011001
a = 4F434152 = 001001111010000110100000101010010
b = D7E58F83 = 011010111111001011000111110000011
c = 68BF5F65 = 001101000101111110101111101100101
d = 352DB6C0 = 000110101001011011011011011000000
e = 73769D64 = 001110011011101101001110101100100
f = DF4E1862 = 011011111010011100001100001100010
g = 71051E01 = 001110001000001010001111000000001
h = 870F00D0 = 010000111000011110000000011010000

Step 7. Change the final values

After the compression cycle, but still, inside the main cycle, we modify the hash values ​​by adding the corresponding variables a... to them h. As usual, all addition is done modulo 2 ^ 32.

h0 = h0 + a = 10111001010011010010011110111001
h1 = h1 + b = 10010011010011010011111000001000
h2 = h2 + c = 10100101001011100101001011010111
h3 = h3 + d = 11011010011111011010101111111010
h4 = h4 + e = 11000100100001001110111111100011
h5 = h5 + f = 01111010010100111000000011101110
h6 = h6 + g = 10010000100010001111011110101100
h7 = h7 + h = 11100010111011111100110111101001

Step 8. Get the final hash

And the last important step is putting everything together.

digest = h0 append h1 append h2 append h3 append h4 append h5 append h6 append h7
= B94D27B9934D3E08A52E52D7DA7DABFAC484EFE37A5380EE9088F7ACE2EFCDE9

Done! We have performed every SHA-2 (SHA-256) step (without some iterations).

SHA-2 pseudocode algorithm

If you want to look at all the steps we just took in pseudocode, here’s an example :

Explanations:
All variables are unsigned, have a size of 32 bits and are summed modulo 232 during calculations
message - the original binary message
m - transformed message
Initializing Variables
(the first 32 bits of the fractional parts of the square roots of the first eight primes [2 to 19]):
h0 := 0x6a09e667
h1 := 0xbb67ae85
h2 := 0x3c6ef372
h3 := 0xa54ff53a
h4 := 0x510e527f
h5 := 0x9b05688c
h6 := 0x1f83d9ab
h7 := 0x5be0cd19
Constant table
(the first 32 bits of the fractional parts of the cube roots of the first 64 primes [2 to 311]):
k[ 0..63 ] :=
0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5, 0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3, 0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc, 0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7, 0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13, 0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3, 0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5, 0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208, 0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2
Preliminary processing:
m: = message ǁ [single bit]
m: = m ǁ [k zero bits], where k is the smallest non-negative number such that
(L + 1 + K) mod 512 = 448, where L is the number of bits in the message (comparable modulo 512 with 448)
m: = m ǁ Length (message) - the length of the original message in bits as a 64-bit number
in byte order from high to low
Then the message is processed in sequential portions of 512 bits:
split the message into chunks of 512 bits
for each piece
split a chunk into 16 words with a length of 32 bits (with the order of bytes from high to low within a word): w [0..15]
Generate additional 48 words:
for i from 16 to 63
s0 := (w[i-15] rightrotate 7) xor (w[i-15] rightrotate 18) xor (w[i-15] rightshift 3)
s1 := (w[i- 2] rightrotate 17) xor (w[i- 2] rightrotate 19) xor (w[i- 2] rightshift 10)
w[i] := w[i-16] + s0 + w[i-7] + s1
Initialization of auxiliary variables:
a := h0
b := h1
c := h2
d := h3
e := h4
f := h5
g := h6
h := h7
Main loop:
for i from 0 to63
S1 := (e rightrotate 6) xor (e rightrotate 11) xor (e rightrotate 25)
ch := (e and f) xor ((not e) and g)
temp1 := h + S1 + ch + k[i] + w[i]
S0 := (a rightrotate 2) xor (a rightrotate 13) xor (a rightrotate 22)
maj := (a and b) xor (a and c) xor (b and c)
temp2 := S0 + maj

h := g
g := f
f := e
e := d + temp1
d := c
c := b
b := a
a := temp1 + temp2
Add the obtained values to the previously calculated result:
h0 := h0 + a
h1 := h1 + b
h2 := h2 + c
h3 := h3 + d
h4 := h4 + e
h5 := h5 + f
h6 := h6 + g
h7 := h7 + h
Get the final SHA-2 hash value:
digest := hash := h0 append h1 append h2 append h3 append h4 append h5 append h6 append h7

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Written by

Bioinformatician at Oncobox Inc. (@oncobox). Research Associate at Moscow Institute of Physics and Technology (@mipt_eng).

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