From 63be5b53b6d15f7706ad21e9801dae5b723e8340 Mon Sep 17 00:00:00 2001 From: Eric Biggers Date: Tue, 14 Feb 2017 13:43:27 -0800 Subject: [PATCH] crypto: gf128mul - fix some comments Fix incorrect references to GF(128) instead of GF(2^128), as these are two entirely different fields, and fix a few other incorrect comments. Cc: Alex Cope Signed-off-by: Eric Biggers Signed-off-by: Herbert Xu --- crypto/gf128mul.c | 13 +++++++------ include/crypto/gf128mul.h | 26 ++++++++++++++------------ 2 files changed, 21 insertions(+), 18 deletions(-) diff --git a/crypto/gf128mul.c b/crypto/gf128mul.c index 72015fee533d..d9e3eecc218a 100644 --- a/crypto/gf128mul.c +++ b/crypto/gf128mul.c @@ -44,7 +44,7 @@ --------------------------------------------------------------------------- Issue 31/01/2006 - This file provides fast multiplication in GF(128) as required by several + This file provides fast multiplication in GF(2^128) as required by several cryptographic authentication modes */ @@ -116,9 +116,10 @@ static const u16 gf128mul_table_lle[256] = gf128mul_dat(xda_lle); static const u16 gf128mul_table_bbe[256] = gf128mul_dat(xda_bbe); -/* These functions multiply a field element by x, by x^4 and by x^8 - * in the polynomial field representation. It uses 32-bit word operations - * to gain speed but compensates for machine endianess and hence works +/* + * The following functions multiply a field element by x or by x^8 in + * the polynomial field representation. They use 64-bit word operations + * to gain speed but compensate for machine endianness and hence work * correctly on both styles of machine. */ @@ -251,7 +252,7 @@ EXPORT_SYMBOL(gf128mul_bbe); /* This version uses 64k bytes of table space. A 16 byte buffer has to be multiplied by a 16 byte key - value in GF(128). If we consider a GF(128) value in + value in GF(2^128). If we consider a GF(2^128) value in the buffer's lowest byte, we can construct a table of the 256 16 byte values that result from the 256 values of this byte. This requires 4096 bytes. But we also @@ -330,7 +331,7 @@ EXPORT_SYMBOL(gf128mul_64k_bbe); /* This version uses 4k bytes of table space. A 16 byte buffer has to be multiplied by a 16 byte key - value in GF(128). If we consider a GF(128) value in a + value in GF(2^128). If we consider a GF(2^128) value in a single byte, we can construct a table of the 256 16 byte values that result from the 256 values of this byte. This requires 4096 bytes. If we take the highest byte in diff --git a/include/crypto/gf128mul.h b/include/crypto/gf128mul.h index 592d47e565a8..9662c4538873 100644 --- a/include/crypto/gf128mul.h +++ b/include/crypto/gf128mul.h @@ -43,7 +43,7 @@ --------------------------------------------------------------------------- Issue Date: 31/01/2006 - An implementation of field multiplication in Galois Field GF(128) + An implementation of field multiplication in Galois Field GF(2^128) */ #ifndef _CRYPTO_GF128MUL_H @@ -65,7 +65,7 @@ * are left and the lsb's are right. char b[16] is an array and b[0] is * the first octet. * - * 80000000 00000000 00000000 00000000 .... 00000000 00000000 00000000 + * 10000000 00000000 00000000 00000000 .... 00000000 00000000 00000000 * b[0] b[1] b[2] b[3] b[13] b[14] b[15] * * Every bit is a coefficient of some power of X. We can store the bits @@ -85,15 +85,17 @@ * Both of the above formats are easy to implement on big-endian * machines. * - * EME (which is patent encumbered) uses the ble format (bits are stored - * in big endian order and the bytes in little endian). The above buffer - * represents X^7 in this case and the primitive polynomial is b[0] = 0x87. + * XTS and EME (the latter of which is patent encumbered) use the ble + * format (bits are stored in big endian order and the bytes in little + * endian). The above buffer represents X^7 in this case and the + * primitive polynomial is b[0] = 0x87. * * The common machine word-size is smaller than 128 bits, so to make * an efficient implementation we must split into machine word sizes. - * This file uses one 32bit for the moment. Machine endianness comes into - * play. The lle format in relation to machine endianness is discussed - * below by the original author of gf128mul Dr Brian Gladman. + * This implementation uses 64-bit words for the moment. Machine + * endianness comes into play. The lle format in relation to machine + * endianness is discussed below by the original author of gf128mul Dr + * Brian Gladman. * * Let's look at the bbe and ble format on a little endian machine. * @@ -127,10 +129,10 @@ * machines this will automatically aligned to wordsize and on a 64-bit * machine also. */ -/* Multiply a GF128 field element by x. Field elements are held in arrays - of bytes in which field bits 8n..8n + 7 are held in byte[n], with lower - indexed bits placed in the more numerically significant bit positions - within bytes. +/* Multiply a GF(2^128) field element by x. Field elements are + held in arrays of bytes in which field bits 8n..8n + 7 are held in + byte[n], with lower indexed bits placed in the more numerically + significant bit positions within bytes. On little endian machines the bit indexes translate into the bit positions within four 32-bit words in the following way