openssl is acting open to any size key

how does openssl works with key as it is taking any size of key (1 byte to any size). What is the procedure to go to actual key here ..

openssl enc -d -des-ecb -in cipher.txt -out text.out -K '530343412312345445123345677812345678812324'

回答1:

how does openssl works with key ... What is the procedure...

It depends on the program, but procedures are usually consistent across the library. In you example, you are using the openssl dec, so you are using the dec sub-program. The source code is available in <openssl dir>/apps/enc.c (enc and dec are part of enc.c).

Here's the relevant parts:

unsigned char key[EVP_MAX_KEY_LENGTH],iv[EVP_MAX_IV_LENGTH];
unsigned char salt[PKCS5_SALT_LEN];
...
char *hkey=NULL,*hiv=NULL,*hsalt = NULL;

The argument to -K is stored in hkey:

else if (strcmp(*argv,"-K") == 0)
{
    if (--argc < 1) goto bad;
    hkey= *(++argv);
}

Then, around line 580:

if ((hkey != NULL) && !set_hex(hkey,key,sizeof key))
{
    /* Handle failure */
}

set_hex is shown below and hex decodes the argument passed in through -K. It back fills the unused length with 0's via the memset. The unused length is EVP_MAX_KEY_LENGTH minus the length -K argument (after hex decoding).

Finally, around line 610:

if (!EVP_CipherInit_ex(ctx, NULL, NULL, key, iv, enc))
{
    /* Handle failure */
}

Note: -k (small k) takes a different code path and uses EVP_BytesToKey to derive the key.

int set_hex(char *in, unsigned char *out, int size)
{
    int i,n;
    unsigned char j;

    n=strlen(in);
    if (n > (size*2))
    {
        BIO_printf(bio_err,"hex string is too long\n");
        return(0);
    }
    memset(out,0,size);
    for (i=0; i<n; i++)
    {
        j=(unsigned char)*in;
        *(in++)='\0';
        if (j == 0) break;
        if ((j >= '0') && (j <= '9'))
            j-='0';
        else if ((j >= 'A') && (j <= 'F'))
            j=j-'A'+10;
        else if ((j >= 'a') && (j <= 'f'))
            j=j-'a'+10;
        else
        {
            BIO_printf(bio_err,"non-hex digit\n");
            return(0);
        }
        if (i&1)
            out[i/2]|=j;
        else
            out[i/2]=(j<<4);
    }
    return(1);
}

回答2:

My observation to the case gave following conclusion: