A Step-By-Step Look into How SHA-256 Works

How SHA-256 Works Step-By-Step SHA-2 (Secure Hash Algorithm 2), of which SHA-256 is a part, is one of the most popular around. A cryptographic hash, also often referred to as a “digest”, “fingerprint” or “signature”, is an string of characters that is generated from a separate piece of input text. For example, SHA-256 generates a 256-bit (32-byte) signature. hash algorithms almost perfectly unique Further down in this article, we will break down each step of SHA 256’s and work through a real example by hand. cryptographic algorithm What is a hash function? Like we mentioned above, a hash function generates a “fingerprint” for a given input string. For example, if we were to hash the entire text of JRR Tolkien’s “The Lord of The Rings” series using the SHA 256 algorithm, we would get a 256-bit output almost unique to that book’s text. If we changed even a single letter in the book, the output hash would be different. wildly It’s worth noting that we say the output of a hash is “almost unique” because there are a number of output strings, after all, there are only ever 256 bits in the output. The number of possible inputs, however, is , meaning some inputs will hash to the same output. When this happens, it’s called a “collision”, and it is nearly impossible. After all, in SHA-256 there are possible outputs. Let me write out that number for you: finite infinite 2^256 115,792,089,237,316,195,423,570,985,008,687,907,853,269,984,665,640,564,039,457,584,007,913,129,639,936 possible outputs for SHA-256 The three of the main purposes of a hash function are: To scramble data deterministically To accept an input of arbitrary length and output a fixed length result To manipulate data irreversibly. The input cannot be derived from the output SHA-2 is a very famous and strong family of hash functions because, as you would expect, it serves all the purposes mentioned above. SHA-2 family vs SHA-256 SHA-2 is an algorithm, a generalized idea of how to hash data. SHA-2 has several variants, all of which use the same algorithm but use different constants. SHA-256, for example, sets additional constants that define the behavior of the SHA-2 algorithm, one of these constants is the output size, 256. The 256 and 512 in SHA-256 and SHA-512 refer to the respective digest size in bits. SHA-2 family vs SHA-1 SHA-2 is a successor to the SHA-1 hash and remains one of the strongest hash functions in use today. SHA-256, as opposed to SHA-1, hasn’t been compromised. For this reason, there’s really no reason to use SHA-1 these days, it isn’t safe. The flexibility of output size (224, 256, 512, etc) also allows SHA-2 to pair well with popular and ciphers like . KDFs AES-256 Formal acceptance by NIST SHA-256 is formally defined in the National Institute of Standards and Technology’s . Along with standardization and formalization comes a list of that allow developers to ensure they’ve implemented the algorithm properly. FIPS 180-4 test vectors SHA-2 is known for its security (it hasn’t ) and its speed. In cases where , such as mining Bitcoin, a fast hash algorithm like SHA-2 often has the upper hand. broken down like SHA-1 keys are not generated Step-by-step SHA-256 hash of “hello world” Step 1 - Pre-Processing Convert “hello world” to binary: 01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111 01110010 01101100 01100100 Append a single 1: 01101000 01100101 01101100 01101100 01101111 00100000 01110111 01101111 01110010 01101100 01100100 1 Pad with 0’s until data is a multiple of 512, less 64 bits (448 bits in our case): 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 Append 64 bits to the end, where the 64 bits are a integer representing the length of the original input in binary. In our case, 88, or in binary, “1011000”. big-endian 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 our input, which will always be evenly divisible by 512. Step 2 - Initialize Hash Values (h) Now we create 8 hash values. These are hard-coded constants that represent 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 - Initialize Round Constants (k) Similar to step 2, we are creating some constants ( ). This time, there are 64 of them. Each value (0-63) is the first 32 bits of the fractional parts of the cube roots of the first 64 primes (2 - 311). Learn more about constants and when to use them here 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 - Chunk Loop The following steps will happen for each 512-bit “chunk” of data from our input. In our case, because “hello world” is so short, we only have one chunk. At each iteration of the loop, we will be mutating the hash values h0-h7, which will be the final output. Step 5 - Create Message Schedule (w) Copy the input data from step 1 into a new array where each entry is a 32-bit word: 01101000011001010110110001101100 01101111001000000111011101101111 01110010011011000110010010000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000000000000 00000000000000000000000001011000 Add 48 more words initialized to zero, such that we have 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 Modify the zero-ed indexes at the end of the array using the following algorithm: For from w[16…63]: i 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 Let’s do w[16] so we can see how it works: w[1] rightrotate 7: 01101111001000000111011101101111 -> 11011110110111100100000011101110 w[1] rightrotate 18: 01101111001000000111011101101111 -> 00011101110110111101101111001000 w[1] rightshift 3: 01101111001000000111011101101111 -> 00001101111001000000111011101101 s0 = 11011110110111100100000011101110 XOR 00011101110110111101101111001000 XOR 00001101111001000000111011101101 s0 = 11001110111000011001010111001011 w[14] rightrotate 17: 00000000000000000000000000000000 -> 00000000000000000000000000000000 w[14] rightrotate19: 00000000000000000000000000000000 -> 00000000000000000000000000000000 w[14] rightshift 10: 00000000000000000000000000000000 -> 00000000000000000000000000000000 s1 = 00000000000000000000000000000000 XOR 00000000000000000000000000000000 XOR 00000000000000000000000000000000 s1 = 00000000000000000000000000000000 w[16] = w[0] + s0 + w[9] + s1 w[16] = 01101000011001010110110001101100 + 11001110111000011001010111001011 + 00000000000000000000000000000000 + 00000000000000000000000000000000 // addition is calculated modulo 2^32 w[16] = 00110111010001110000001000110111 This leaves us with 64 words in our message schedule (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 Initialize variables and set them equal to the current hash values respectively. a, b, c, d, e, f, g, h h0, h1, h2, h3, h4, h5, h6, h7 Run the compression loop. The compression loop will mutate the values of . The compression loop is as follows: a…h 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 f = e e = d + temp1 d = c c = b b = a a = temp1 + temp2 Let’s go through the first iteration, all 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 + 01000010100010100010111110011000 + 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 That entire calculation is done 63 more times, modifying the variables a-h throughout. We won’t do it by hand but we would have ender with: 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 = 01001111010000110100000101010010 b = D7E58F83 = 11010111111001011000111110000011 c = 68BF5F65 = 01101000101111110101111101100101 d = 352DB6C0 = 00110101001011011011011011000000 e = 73769D64 = 01110011011101101001110101100100 f = DF4E1862 = 11011111010011100001100001100010 g = 71051E01 = 01110001000001010001111000000001 h = 870F00D0 = 10000111000011110000000011010000 Step 7 - Modify Final Values After the compression loop, but still, within the loop, we modify the hash values by adding their respective variables to them, a-h. As usual, all addition is modulo 2^32. chunk 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 - Concatenate Final Hash Last but not least, slap them all together, a simple will do. string concatenation digest = h0 append h1 append h2 append h3 append h4 append h5 append h6 append h7 = B94D27B9934D3E08A52E52D7DA7DABFAC484EFE37A5380EE9088F7ACE2EFCDE9 Done! We’ve been through every step (sans some iterations) of SHA-256 in excruciating detail :) I’m glad you’ve made it this far! Going step-by-step through the SHA-256 algorithm isn’t exactly a walk in the park. Learning the fundamentals that underpin web security can be a huge boon to your , however, so keep it up! career as a computer scientist The Pseudocode If you want to see all the steps we just did above in pseudocode form, then here is it is, straight from : WikiPedia Note 1: All variables are 32 bit unsigned integers and addition is calculated modulo 232 Note 2: For each round, there is one round constant k[i] and one entry in the message schedule array w[i], 0 ≤ i ≤ 63 Note 3: The compression function uses 8 working variables, a through h Note 4: Big-endian convention is used when expressing the constants in this pseudocode, and when parsing message block data from bytes to words, for example, the first word of the input message "abc" after padding is 0x61626380 Initialize hash values: (first 32 bits of the fractional parts of the square roots of the first 8 primes 2..19): h0 := 0x6a09e667 h1 := 0xbb67ae85 h2 := 0x3c6ef372 h3 := 0xa54ff53a h4 := 0x510e527f h5 := 0x9b05688c h6 := 0x1f83d9ab h7 := 0x5be0cd19 Initialize array of round constants: (first 32 bits of the fractional parts of the cube roots of the first 64 primes 2..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 Pre-processing (Padding): begin with the original message of length L bits append a single '1' bit append K '0' bits, where K is the minimum number >= 0 such that L + 1 + K + 64 is a multiple of 512 append L as a 64-bit big-endian integer, making the total post-processed length a multiple of 512 bits Process the message in successive 512-bit chunks: break message into 512-bit chunks for each chunk create a 64-entry message schedule array w[0..63] of 32-bit words (The initial values in w[0..63] don't matter, so many implementations zero them here) copy chunk into first 16 words w[0..15] of the message schedule array Extend the first 16 words into the remaining 48 words w[16..63] of the message schedule array: 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 Initialize working variables to current hash value: a := h0 b := h1 c := h2 d := h3 e := h4 f := h5 g := h6 h := h7 Compression function main loop: 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 f := e e := d + temp1 d := c c := b b := a a := temp1 + temp2 Add the compressed chunk to the current hash value: h0 := h0 + a h1 := h1 + b h2 := h2 + c h3 := h3 + d h4 := h4 + e h5 := h5 + f h6 := h6 + g h7 := h7 + h Produce the final hash value (big-endian): digest := hash := h0 append h1 append h2 append h3 append h4 append h5 append h6 append h7 Also Published Here