这篇教程C++ BN_lshift1函数代码示例写得很实用，希望能帮到您。
本文整理汇总了C++中BN_lshift1函数的典型用法代码示例。如果您正苦于以下问题：C++ BN_lshift1函数的具体用法？C++ BN_lshift1怎么用？C++ BN_lshift1使用的例子？那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。 在下文中一共展示了BN_lshift1函数的17个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于我们的系统推荐出更棒的C++代码示例。 示例1: BN_mod_lshift1_quick/* BN_mod_lshift1 variant that may be used if a is non-negative * and less than m */int BN_mod_lshift1_quick(BIGNUM *r, const BIGNUM *a, const BIGNUM *m) { if (!BN_lshift1(r, a)) return 0; if (BN_cmp(r, m) >= 0) return BN_sub(r, r, m); return 1; }
开发者ID:12019，项目名称:svn.gov.pt，代码行数:9，
示例2: BN_divint BN_div(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m, const BIGNUM *d, BN_CTX *ctx) { int i,nm,nd; int ret = 0; BIGNUM *D; bn_check_top(m); bn_check_top(d); if (BN_is_zero(d)) { BNerr(BN_F_BN_DIV,BN_R_DIV_BY_ZERO); return(0); } if (BN_ucmp(m,d) < 0) { if (rem != NULL) { if (BN_copy(rem,m) == NULL) return(0); } if (dv != NULL) BN_zero(dv); return(1); } BN_CTX_start(ctx); D = BN_CTX_get(ctx); if (dv == NULL) dv = BN_CTX_get(ctx); if (rem == NULL) rem = BN_CTX_get(ctx); if (D == NULL || dv == NULL || rem == NULL) goto end; nd=BN_num_bits(d); nm=BN_num_bits(m); if (BN_copy(D,d) == NULL) goto end; if (BN_copy(rem,m) == NULL) goto end; /* The next 2 are needed so we can do a dv->d[0]|=1 later * since BN_lshift1 will only work once there is a value :-) */ BN_zero(dv); if(bn_wexpand(dv,1) == NULL) goto end; dv->top=1; if (!BN_lshift(D,D,nm-nd)) goto end; for (i=nm-nd; i>=0; i--) { if (!BN_lshift1(dv,dv)) goto end; if (BN_ucmp(rem,D) >= 0) { dv->d[0]|=1; if (!BN_usub(rem,rem,D)) goto end; }/* CAN IMPROVE (and have now :=) */ if (!BN_rshift1(D,D)) goto end; } rem->neg=BN_is_zero(rem)?0:m->neg; dv->neg=m->neg^d->neg; ret = 1; end: BN_CTX_end(ctx); return(ret); }
开发者ID:EddieGarmon，项目名称:netduino-netmf，代码行数:60，
示例3: BN_mod_lshift1int BN_mod_lshift1(BIGNUM *r, const BIGNUM *a, const BIGNUM *m, BN_CTX *ctx){ if (!BN_lshift1(r, a)) return 0; bn_check_top(r); return BN_nnmod(r, r, m, ctx);}
开发者ID:1234-，项目名称:openssl，代码行数:7，
示例4: one/* The secret integers s0 and s1 must be in the range 0 < s < n for some n, and must be relatively prime to that n. We know a priori that n is of the form 2**k * p for some small integer k and prime p. Therefore, it suffices to choose a random integer in the range [0, n/2), multiply by two and add one (enforcing oddness), and then reject values which are divisible by p. */static BIGNUM *random_s(const BIGNUM *n, const BIGNUM *p, BN_CTX *c){ BIGNUM h, m, *r; BN_init(&h); BN_init(&m); FAILZ(r = BN_new()); FAILZ(BN_copy(&h, n)); FAILZ(BN_rshift1(&h, &h)); do { FAILZ(BN_rand_range(r, &h)); FAILZ(BN_lshift1(r, r)); FAILZ(BN_add(r, r, BN_value_one())); FAILZ(BN_nnmod(&m, r, p, c)); } while (BN_is_zero(&m)); BN_clear(&h); BN_clear(&m); return r; fail: BN_clear(&h); BN_clear(&m); if (r) BN_clear_free(r); return 0;}
开发者ID:zackw，项目名称:moeller-ref，代码行数:34，
示例5: fermat_question_askstatic RSA *fermat_question_ask(const RSA *rsa){ BIGNUM *a = BN_new(), *b = BN_new(), *a2 = BN_new(), *b2 = BN_new(); BIGNUM *n = rsa->n; BIGNUM *tmp = BN_new(), *rem = BN_new(), *dssdelta = BN_new(); BN_CTX *ctx = BN_CTX_new(); RSA *ret = NULL; BN_sqrtmod(tmp, rem, n, ctx); /* Δ = |p - q| = |a + b - a + b| = |2b| > √N 2
C++ BN_mod函数代码示例
C++ BN_lshift函数代码示例
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