Making conversion similar to BN_hex2bn + BN_bn2bin without openSSL in C

Content Index :

Making conversion similar to BN_hex2bn + BN_bn2bin without openSSL in C
Trouble with OpenSSL's BN_bn2bin function
equivalent to BN_hex2bn
Unsure of parameter to store BIGNUM in char - BN_bn2bin()
Use of OpenSSL BN_hex2bn in Swift 4 - How to build the handle
BN_hex2bn magically segfaults in openSSL

Making conversion similar to BN_hex2bn + BN_bn2bin without openSSL in C

Tag : c , By : dlouzan

Date : November 29 2020, 04:01 AM

will help you
BN_bn2bin "creates a representation that is truly binary (i.e. a sequence of bits). More specifically, it creates a big-endian representation of the number." So this is not just an array of 0s and 1s, right?

#include <stdio.h>
#include <openssl/bn.h>

int main(
    int argc,
    char **argv)
{
    const char *hexString = argv[1];

    BIGNUM *bnEncr = BN_new();
    BN_hex2bn(&bnEncr, hexString);
    unsigned int numOfBytesEncr = BN_num_bytes(bnEncr);
    unsigned char encrBin[numOfBytesEncr];
    BN_bn2bin(bnEncr, encrBin);
    fwrite(encrBin, 1, numOfBytesEncr, stdout);
}

$ ./bntest 74162ac74759e85654e0e7762c2cdd26 | hexdump -C
00000000  74 16 2a c7 47 59 e8 56  54 e0 e7 76 2c 2c dd 26 |t.*.GY.VT..v,,.&|
00000010

$ ./bntest  74162ac74759e85654e0e7762c2cdd26 | xxd -b -c 4
00000000: 01110100 00010110 00101010 11000111  t.*.
00000004: 01000111 01011001 11101000 01010110  GY.V
00000008: 01010100 11100000 11100111 01110110  T..v
0000000c: 00101100 00101100 11011101 00100110  ,,.&

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Making conversion similar to BN_hex2bn + BN_bn2bin without openSSL in C , Trouble with OpenSSLs BN_bn2bin function , BN_hex2bn magically segfaults in openSSL , Use of OpenSSL BN_hex2bn in Swift 4 - How to build the handle , what is openssl ,

Trouble with OpenSSL's BN_bn2bin function

Tag : c , By : koder

Date : March 29 2020, 07:55 AM

may help you . BN_bn2bin doesn't create a printable string - instead, it creates a representation that is truly binary (i.e. a sequence of bits). More specifically, it createas a big-endian representation of the number. Since 42 fits into one byte, you get one byte 0x2a, which is "*" in ASCII.
If you want a 0/1 representation, you need to iterate over all bytes, and do the printing yourself (e.g. with shifting or a lookup table).

equivalent to BN_hex2bn

Tag : chash , By : kameel

Date : March 29 2020, 07:55 AM

Does that help is there any equivalent to BN_hex2bn in C# , I wanna make BigInteger from hex string!

0894B645E89E1535BBDAD5B8B290650530801B18EBFBF5E8FAB3C82872A3E9BB7

62100066509156017342069496140902949863249758336000796928566441170293728648119

Unsure of parameter to store BIGNUM in char - BN_bn2bin()

Tag : c , By : B3CFT

Date : March 29 2020, 07:55 AM

seems to work fine len is the length of the buffer pointed to by p. The function pads out the converted bignum to len bytes (with zeroes, which doesn't change the mathematical value of the number represented).
That function appears to have a bug, by the way - I strongly suspect that the memset() call should be memset(p, 0, pad);.

Use of OpenSSL BN_hex2bn in Swift 4 - How to build the handle

Tag : swift , By : Tink

Date : March 29 2020, 07:55 AM

To fix the issue you can do BN_hex2bn() expects the address of a BIGNUM * as the first argument, in Swift that is the address of an UnsafeMutablePointer? variable. You “only” have to declare that variable (optionals are automatically initialized with nil) and pass it as an inout-argument with & to the function. Example:

let hexString = "123456789abcdef"

var bn: UnsafeMutablePointer<BIGNUM>?
BN_hex2bn(&bn, hexString)

// Do something with `bn` ...   

// Eventually free the memory:
BN_free(bn)

class BN {
    var bn: UnsafeMutablePointer<BIGNUM>?

    init(hexString: String) {
        BN_hex2bn(&bn, hexString)
    }

    deinit {
        if let bn = bn { BN_free(bn) }
    }
}