field_impl.h File Reference

#include "field.h"
#include "util.h"
#include "field_10x26_impl.h"

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Functions
static SECP256K1_INLINE void	secp256k1_fe_clear (secp256k1_fe *a)
static SECP256K1_INLINE int	secp256k1_fe_equal (const secp256k1_fe *a, const secp256k1_fe *b)
static int	secp256k1_fe_sqrt (secp256k1_fe *SECP256K1_RESTRICT r, const secp256k1_fe *SECP256K1_RESTRICT a)
static void	secp256k1_fe_verify (const secp256k1_fe *a)
static void	secp256k1_fe_verify_magnitude (const secp256k1_fe *a, int m)

Function Documentation

secp256k1_fe_clear()

static SECP256K1_INLINE void secp256k1_fe_clear ( secp256k1_fe *  a )

static

Definition at line 21 of file field_impl.h.

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secp256k1_fe_equal()

static SECP256K1_INLINE int secp256k1_fe_equal ( const secp256k1_fe *  a, const secp256k1_fe *  b  )

static

Definition at line 25 of file field_impl.h.

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secp256k1_fe_sqrt()

static int secp256k1_fe_sqrt ( secp256k1_fe *SECP256K1_RESTRICT  r, const secp256k1_fe *SECP256K1_RESTRICT  a  )

static

Given that p is congruent to 3 mod 4, we can compute the square root of a mod p as the (p+1)/4'th power of a.

As (p+1)/4 is an even number, it will have the same result for a and for (-a). Only one of these two numbers actually has a square root however, so we test at the end by squaring and comparing to the input. Also because (p+1)/4 is an even number, the computed square root is itself always a square (a ** ((p+1)/4) is the square of a ** ((p+1)/8)).

The binary representation of (p + 1)/4 has 3 blocks of 1s, with lengths in { 2, 22, 223 }. Use an addition chain to calculate 2^n - 1 for each block: 1, [2], 3, 6, 9, 11, [22], 44, 88, 176, 220, [223]

Definition at line 37 of file field_impl.h.

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secp256k1_fe_verify()

static void secp256k1_fe_verify ( const secp256k1_fe *  a )

static

Definition at line 149 of file field_impl.h.

secp256k1_fe_verify_magnitude()

static void secp256k1_fe_verify_magnitude ( const secp256k1_fe *  a, int  m  )

static

Definition at line 150 of file field_impl.h.