谨慎使用BN_bin2bn和BN_bn2bin

最新推荐文章于 2024-09-15 04:09:14 发布

原创最新推荐文章于 2024-09-15 04:09:14 发布 · 1.5w 阅读

2 ·

CC 4.0 BY-SA版权

linux 专栏收录该内容

61 篇文章

订阅专栏

本文详细解析了BN_bn2bin函数的使用方法及其特点，通过示例代码展示了该函数如何处理不同大小的大数（BIGNUM），并解释了转换过程中高位零被去除的现象。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

        BIGNUM *a = NULL;
        BIGNUM *b = NULL;
	BIGNUM *c = NULL;
	unsigned char *xychar= NULL;
	int i;
	unsigned char data[20];
		
	const unsigned char testk[20] =
	{
    	0x00, 0x01, 0x43, 0xbd, 0x74, 0x9d, 0xd4, 0x99, 0x00, 0xdb, 0xaa, 0xa1, 0xeb, 0xd2, 0xc9, 0xcb, 0xed, 0x2e, 0xbe, 0x0c
	};
		
	a = BN_new();
        b = BN_new();
	c = BN_new();

	BN_one( a );
	BN_bin2bn( testk, 9, b );
	BN_bin2bn( testk, 20, c );

	//a
	printf("one ->dmax = %d, a:\n", a->dmax);
	xychar = (unsigned char *)a->d;
    for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", xychar[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	memset(data, 0, 20);
	BN_bn2bin( a, data );
	printf("a data:\n");
    for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", data[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	printf("a data BN_num_bits:%d\n", BN_num_bits(a));
	
	//b
	printf("b ->dmax = %d, b:\n", b->dmax);
	xychar = (unsigned char *)b->d;
    for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", xychar[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	memset(data, 0, 20);
	BN_bn2bin( b, data );
	printf("b data:\n");
	for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", data[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	printf("b data BN_num_bits:%d\n", BN_num_bits(b));
	
	//c
	printf("c ->dmax = %d, c:\n", c->dmax);
	xychar = (unsigned char *)c->d;
    for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", xychar[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	memset(data, 0, 20);
	BN_bn2bin( c, data );
	printf("c data:\n");
	for (i = 0; i < 20; i ++ )
    {
        printf( "%02x, ", data[i] );
        if ( ( i + 1 ) % 10 == 0 )
        {
            printf( "\n" );
        }
    }
	printf("c data BN_num_bits:%d\n", BN_num_bits(c));

运算上面code输出如下：

one ->dmax = 1, a:
01, 00, 00, 00, 00, 00, 00, 00, 00, 00,
00, 00, 11, 00, 00, 00, 00, 99, d4, 9d,
a data:
01, 00, 00, 00, 00, 00, 00, 00, 00, 00,
00, 00, 00, 00, 00, 00, 00, 00, 00, 00,
a data BN_num_bits:1
b ->dmax = 3, b:
00, 99, d4, 9d, 74, bd, 43, 01, 00, 00,
00, 00, 19, 00, 00, 00, 0c, be, 2e, ed,
b data:
01, 43, bd, 74, 9d, d4, 99, 00, 00, 00,
00, 00, 00, 00, 00, 00, 00, 00, 00, 00,
b data BN_num_bits:57
c ->dmax = 5, c:
0c, be, 2e, ed, cb, c9, d2, eb, a1, aa,
db, 00, 99, d4, 9d, 74, bd, 43, 01, 00,
c data:
01, 43, bd, 74, 9d, d4, 99, 00, db, aa,
a1, eb, d2, c9, cb, ed, 2e, be, 0c, 00,
c data BN_num_bits:145
one ->dmax = 1, a:
01, 00, 00, 00, 00, 00, 00, 00, 00, 00,
00, 00, 11, 00, 00, 00, 00, 99, d4, 9d,
a data:
01, 00, 00, 00, 00, 00, 00, 00, 00, 00,
00, 00, 00, 00, 00, 00, 00, 00, 00, 00,
a data BN_num_bits:1
b ->dmax = 3, b:
00, 99, d4, 9d, 74, bd, 43, 01, 00, 00,
00, 00, 19, 00, 00, 00, 0c, be, 2e, ed,
b data:
01, 43, bd, 74, 9d, d4, 99, 00, 00, 00,
00, 00, 00, 00, 00, 00, 00, 00, 00, 00,
b data BN_num_bits:57
c ->dmax = 5, c:
0c, be, 2e, ed, cb, c9, d2, eb, a1, aa,
db, 00, 99, d4, 9d, 74, bd, 43, 01, 00,
c data:
01, 43, bd, 74, 9d, d4, 99, 00, db, aa,
a1, eb, d2, c9, cb, ed, 2e, be, 0c, 00,
c data BN_num_bits:145