小端环境下C语言中的SHA-1

时间:2018-11-23 04:34:17

标签: c endianness sha

我碰到的大多数SHA-1实现(甚至在Wikipedia上)都针对大端运行时进行了编码。因此,我正在尝试为自己的机器实现自己的版本(小端)。

我遵循了伪代码形式Wikipedia,并具有以下代码。我找到了一个可转换字节顺序但仍未获得正确输出的函数。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>

#define rotateleft(x,n) ((x<<n) | (x>>(32-n)))

unsigned int endian_reverse(unsigned int n)
{
    unsigned int m = 0;
    m |= n << 24;
    m |= ((n >> 8) << 24) >> 8;
    m |= ((n << 8) >> 24) << 8;
    m |= n >> 24;
    return m;
}

void SHA1(unsigned char * str1)
{
    unsigned long int h0,h1,h2,h3,h4,a,b,c,d,e,f,k,temp;

    h0 = 0x67452301;
    h1 = 0xEFCDAB89;
    h2 = 0x98BADCFE;
    h3 = 0x10325476;
    h4 = 0xC3D2E1F0;

    unsigned char * str;

    str = (unsigned char *)malloc(strlen((const char *)str1)+100);
    strcpy((char *)str,(const char *)str1);

    int current_length = strlen((const char *)str);
    int original_length = current_length;

    str[current_length] = 0x80;
    str[current_length + 1] = '\0';

    char ic = str[current_length];

    current_length++;
    int ib = current_length % 64;

    int i, j;

    if(ib<56)
        ib = 56-ib;
    else
        ib = 120 - ib;

    for(i=0;i < ib;i++)
    {
        str[current_length]=0x00;
        current_length++;
    }

    str[current_length + 1]='\0';

    for(i=0;i<6;i++)
    {
        str[current_length]=0x0;
        current_length++;
    }

    str[current_length] = (original_length * 8) / 0x100 ;
    current_length++;
    str[current_length] = (original_length * 8) % 0x100;
    current_length++;
    str[current_length+i]='\0';

    int number_of_chunks = current_length/64;
    unsigned long int word[80];

    for(i=0;i<number_of_chunks;i++)
    {
        for(j=0;j<16;j++)
        {
            word[j] = 
                str[i*64 + j*4 + 0] * 0x1000000 + str[i*64 + j*4 + 1] * 0x10000 + 
                str[i*64 + j*4 + 2] * 0x100 + str[i*64 + j*4 + 3];
        }
        for(j=16;j<80;j++)
        {
            word[j] = rotateleft((word[j-3] ^ word[j-8] ^ word[j-14] ^ word[j-16]),1);
        }
        a = h0;
        b = h1;
        c = h2;
        d = h3;
        e = h4;
        for(int m=0;m<80;m++)
        {
            if(m<=19)
            {
                f = (b & c) | ((~b) & d);
                k = 0x5A827999;
            }
            else if(m<=39)
            {
                f = b ^ c ^ d;
                k = 0x6ED9EBA1;
            }
            else if(m<=59)
            {
                f = (b & c) | (b & d) | (c & d);
                k = 0x8F1BBCDC;
            }
            else
            {
                f = b ^ c ^ d;
                k = 0xCA62C1D6;
            }
            temp = (rotateleft(a,5) + f + e + k + word[m]) & 0xFFFFFFFF;
            e = d;
            d = c;
            c = rotateleft(b,30);
            b = a;
            a = temp;

        }

        h0 = h0 + a;
        h1 = h1 + b;
        h2 = h2 + c;
        h3 = h3 + d;
        h4 = h4 + e;

    }

    h0 = endian_reverse(h0);
    h1 = endian_reverse(h1);
    h2 = endian_reverse(h2);
    h3 = endian_reverse(h3);
    h4 = endian_reverse(h4);

    printf("\n\n");
    printf("Hash: %x %x %x %x %x",h0, h1, h2, h3, h4);
    printf("\n\n");
}

int main()
{
    SHA1((unsigned char *)"abc");
    return 0;
}

SHA-1(“ abc”):

结果

Hash: f7370736 e388302f 15815610 1ccacd49 e7649bb6

正确(实际)

Hash: a9993e36 4706816a ba3e2571 7850c26c 9cd0d89d

我是否正确或在正确的位置进行字节顺序转换?

1 个答案:

答案 0 :(得分:1)

  

我是否正确或在正确的位置进行字节顺序转换?

endian_reverse(()不是32位时,字节序转换unsigned不正确。

未在正确的位置使用字节顺序转换。不需要字节序转换。

存在其他问题。


代码是假设 unsigned long int是32位。当unsigned long int是64位时,我可以获得与OP相同的答案。

节省时间:在这里没有理由使用松散类型。使用uint32_t, uint8_t。对于数组大小和字符串长度,请使用size_t。避免使用带符号的类型和常量。


通过更改unsigned long-> uint32_t并删除h0 = endian_reverse(h0); ... h7 = endian_reverse(h7);我想到了

Hash: a9993e36 4706816a ba3e2571 7850c26c 9cd0d89d

其他建议。

64的倍数

size中的

str = malloc(size)应该是64的倍数

保持在64的倍数之内

str[current_length+i]='\0';可以写外部分配。

备用尺寸存储

适用于 all size_t个值,最多2 64-3 -1。

  size_t current_length = ...

  // append length in bits
  uint64_t current_length_bits = current_length;
  current_length_bits *= 8;
  for (i = 8; i > 0; ) {
    i--;
    str[current_length + i] = (unsigned char) current_length_bits;
    current_length_bits >>= 8;
  }
  current_length += 8;