openssl is acting open to any size key - cryptography
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'
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);
}
My observation to the case gave following conclusion:
It takes hex value
If the size is less then 8 bytes it pads 0
It takes first 8 bytes as key
Related
How to receive strings from HC05 Bluetooth module using ATmega16 microcontroller
I am having problem in receiving string from HC05 to ATmega16. I am able receive characters but not able to receive strings.
I want to control DC motor wirelessly using ATmega16 and Bluetooth module (HC05). I am sending the timer OCR1A values from serial monitor app to ATmega16 by HC05 but not succeeded.
#define F_CPU 16000000UL
#include<string.h>
#include <avr/io.h>
#include <util/delay.h>
#include <stdlib.h>
#include <stdio.h>
void UART_init()
{
UCSRB |= (1 << RXEN) | (1 << TXEN);
UCSRC |= (1 << URSEL) | (1 << UCSZ0) | (1 << UCS Z1);
UBRRL = 0x67;
}
unsigned char UART_RxChar()
{
while( (UCSRA & (1 << RXC)) == 0 );
return(UDR);
}
void UART_TxChar( char ch )
{
while( !(UCSRA & (1 << UDRE)) ); /* Wait for empty transmit buffer*/
UDR = ch ;
}
void UART_SendString( char* str )
{
unsigned char j = 0;
while( j <= 2 )
{
UART_TxChar( str[j] );
j++;
}
}
int main( void )
{
char buff[3];
char j;
int i = 0, k = 0;
DDRD = (1 << PD5);
UART_init();
while( 1 )
{
buff[0] = UART_RxChar();
buff[1] = UART_RxChar();
buff[2] = UART_RxChar();
j = UART_RxChar();
if( j == '!' )
{
UART_SendString( buff ); // this is to check whether the atmega16 received correct values for timer or not.
UART_SendString( "\n" );
}
}
}
The expected result is when I enter the number in serial monitor app, I should get back the same number on serial monitor app.
In the actual result I am getting different characters sometimes and empty some times.
The string buff is unterminated, so UART_SendString( buff ); will send whatever junk follows the received three characters until a NUL (0) byte is found.
char buff[4] = {0};
Will have room for the NUL and the initialisation will ensure that buff[3] is a NUL terminator.
Alternatively, send the three characters individually since without the terminator they do not constitute a valid C (ASCIIZ) string.
Apart from the lack of nul termination, you code requires input of exactly the form nnn!nnn!nnn!.... If the other end is in fact sending lines with CR or CR+LF terminators - nnn!<newline>nnn!<newline>nnn!<newline>... your receive loop will get out of sync.
A safer solution is to use the previously received three characters whenever a '!' character is received. This can be done in a number of ways - for long buffers a ring-buffer would be advised, but for just three characters it is probably efficient enough to simply shift characters left when inserting a new character - for example:
char buff[4] ;
for(;;)
{
memset( buff, '0', sizeof(buff) - 1 ) ;
char ch = 0 ;
while( (ch != '!' )
{
ch = UART_RxChar() ;
if( isdigit(ch) )
{
// Shift left one digit
memmove( buff, &buff[1], sizeof(buff) - 2 ) ;
// Insert new digit at the right
buff[sizeof(buff) - 2] = ch ;
}
else if( ch != '!' )
{
// Unexpected character, reset buffer
memset( buff, '0', sizeof(buff) - 1 ) ;
}
}
UART_SendString( buff ) ;
UART_SendString( "\n" ) ;
}
This also has the advantage that it will work when the number entered is less than three digits, and will discard any sequence containing non-digit characters.
What does PKCS5_PBKDF2_HMAC_SHA1 return value mean?
I'm attempting to use OpenSSL's PKCS5_PBKDF2_HMAC_SHA1 method. I gather that it returns 0 if it succeeds, and some other value otherwise. My question is, what does a non-zero return value mean? Memory error? Usage error? How should my program handle it (retry, quit?)? Edit: A corollary question is, is there any way to figure this out besides reverse-engineering the method itself?
is there any way to figure this out besides reverse-engineering the method itself?
PKCS5_PBKDF2_HMAC_SHA1 looks like one of those undocumented functions because I can't find it in the OpenSSL docs. OpenSSL has a lot of them, so you should be prepared to study the sources if you are going to use the library.
I gather that it returns 0 if it succeeds, and some other value otherwise.
Actually, its reversed. Here's how I know...
$ grep -R PKCS5_PBKDF2_HMAC_SHA1 *
crypto/evp/evp.h:int PKCS5_PBKDF2_HMAC_SHA1(const char *pass, int passlen,
crypto/evp/p5_crpt2.c:int PKCS5_PBKDF2_HMAC_SHA1(const char *pass, int passlen,
...
So, you find the function's implementation in crypto/evp/p5_crpt2.c:
int PKCS5_PBKDF2_HMAC_SHA1(const char *pass, int passlen,
const unsigned char *salt, int saltlen, int iter,
int keylen, unsigned char *out)
{
return PKCS5_PBKDF2_HMAC(pass, passlen, salt, saltlen, iter,
EVP_sha1(), keylen, out);
}
Following PKCS5_PBKDF2_HMAC:
$ grep -R PKCS5_PBKDF2_HMAC *
...
crypto/evp/evp.h:int PKCS5_PBKDF2_HMAC(const char *pass, int passlen,
crypto/evp/p5_crpt2.c:int PKCS5_PBKDF2_HMAC(const char *pass, int passlen,
...
And again, from crypto/evp/p5_crpt2.c:
int PKCS5_PBKDF2_HMAC(const char *pass, int passlen,
const unsigned char *salt, int saltlen, int iter,
const EVP_MD *digest,
int keylen, unsigned char *out)
{
unsigned char digtmp[EVP_MAX_MD_SIZE], *p, itmp[4];
int cplen, j, k, tkeylen, mdlen;
unsigned long i = 1;
HMAC_CTX hctx_tpl, hctx;
mdlen = EVP_MD_size(digest);
if (mdlen < 0)
return 0;
HMAC_CTX_init(&hctx_tpl);
p = out;
tkeylen = keylen;
if(!pass)
passlen = 0;
else if(passlen == -1)
passlen = strlen(pass);
if (!HMAC_Init_ex(&hctx_tpl, pass, passlen, digest, NULL))
{
HMAC_CTX_cleanup(&hctx_tpl);
return 0;
}
while(tkeylen)
{
if(tkeylen > mdlen)
cplen = mdlen;
else
cplen = tkeylen;
/* We are unlikely to ever use more than 256 blocks (5120 bits!)
* but just in case...
*/
itmp[0] = (unsigned char)((i >> 24) & 0xff);
itmp[1] = (unsigned char)((i >> 16) & 0xff);
itmp[2] = (unsigned char)((i >> 8) & 0xff);
itmp[3] = (unsigned char)(i & 0xff);
if (!HMAC_CTX_copy(&hctx, &hctx_tpl))
{
HMAC_CTX_cleanup(&hctx_tpl);
return 0;
}
if (!HMAC_Update(&hctx, salt, saltlen)
|| !HMAC_Update(&hctx, itmp, 4)
|| !HMAC_Final(&hctx, digtmp, NULL))
{
HMAC_CTX_cleanup(&hctx_tpl);
HMAC_CTX_cleanup(&hctx);
return 0;
}
HMAC_CTX_cleanup(&hctx);
memcpy(p, digtmp, cplen);
for(j = 1; j < iter; j++)
{
if (!HMAC_CTX_copy(&hctx, &hctx_tpl))
{
HMAC_CTX_cleanup(&hctx_tpl);
return 0;
}
if (!HMAC_Update(&hctx, digtmp, mdlen)
|| !HMAC_Final(&hctx, digtmp, NULL))
{
HMAC_CTX_cleanup(&hctx_tpl);
HMAC_CTX_cleanup(&hctx);
return 0;
}
HMAC_CTX_cleanup(&hctx);
for(k = 0; k < cplen; k++)
p[k] ^= digtmp[k];
}
tkeylen-= cplen;
i++;
p+= cplen;
}
HMAC_CTX_cleanup(&hctx_tpl);
return 1;
}
So it looks like 0 on failure, and 1 on success. You should not see other values. And if you get a 0, then all the OUT parameters are junk.
Memory error? Usage error?
Well, sometimes you can call ERR_get_error. If you call it and it makes sense, then the error code is good. If the error code makes no sense, then its probably not good.
Sadly, that's the way I handle it because the library is not consistent with setting error codes. For example, here's the library code to load the RDRAND engine.
Notice the code clears the error code on failure if its a 3rd generation Ivy Bridge (that's the capability being tested), and does not clear or set an error otherwise!!!
void ENGINE_load_rdrand (void)
{
extern unsigned int OPENSSL_ia32cap_P[];
if (OPENSSL_ia32cap_P[1] & (1<<(62-32)))
{
ENGINE *toadd = ENGINE_rdrand();
if(!toadd) return;
ENGINE_add(toadd);
ENGINE_free(toadd);
ERR_clear_error();
}
}
How should my program handle it (retry, quit?)?
It looks like a hard failure.
Finally, that's exactly how I navigate the sources in this situation. If you don't like grep you can try ctags or another source code browser.
What's the use for the username and hostname at the endof the public RSA key
What is the purpose of: username#hostname at the end of the RSA public key? I know that it matches the the generator of the key, but is it ever used for anything significant?
It is only a comment, to help you keep straight where each public key comes from.
In the openSSH source (v6.3,auth-rsa.c:57-65):
/*
* The .ssh/authorized_keys file contains public keys, one per line, in the
* following format:
* options bits e n comment
* where bits, e and n are decimal numbers,
* and comment is any string of characters up to newline. The maximum
* length of a line is SSH_MAX_PUBKEY_BYTES characters. See sshd(8) for a
* description of the options.
*/
And reading:
case KEY_RSA1:
/* Get number of bits. */
if (*cp < '0' || *cp > '9')
return -1; /* Bad bit count... */
for (bits = 0; *cp >= '0' && *cp <= '9'; cp++)
bits = 10 * bits + *cp - '0';
if (bits == 0)
return -1;
*cpp = cp;
/* Get public exponent, public modulus. */
if (!read_bignum(cpp, ret->rsa->e))
return -1;
if (!read_bignum(cpp, ret->rsa->n))
return -1;
/* validate the claimed number of bits */
if ((u_int)BN_num_bits(ret->rsa->n) != bits) {
verbose("key_read: claimed key size %d does not match "
"actual %d", bits, BN_num_bits(ret->rsa->n));
return -1;
}
success = 1;
break;
It doesn't even parse the comment.
How to validate an ASCII string with a two digit hexadecimal checksum appended?
I am using a Renesas 16 bt MCU with HEW (High-performance Embedded Workbench) compiler.
The system receives ACSII data of the form:
<data><cc>
where <cc> comprises two ASCII hex digits corresponding to the 8-bit bitwise XOR of all the preceding characters. The maximum length of the string including <cc> is 14.
Here is my attempt:
#pragma INTERRUPT Interrupt_Rx0
void Interrupt_Rx0 (void)
{
unsigned char rx_byte, rx_status_byte,hex;
char buffer[15],test[5];
int r,k[15];
char * pEnd;
unsigned char dat,arr[14],P3;
unsigned int i,P1[10];
rx_byte = u0rbl; //get rx data
rx_status_byte = u0rbh;
if ((rx_status_byte & 0x80) == 0x00) //if no error
{
if ((bf_rx0_start == 0) && (rx_byte == '?') && (bf_rx0_ready == 0))
{
byte_rx0_buffer[0]=rx_byte;
bf_rx0_start = 1;
byte_rx0_ptr = 1;
}
else
{
if (rx_byte == '?')
{
bf_rx0_start = 1;
byte_rx0_ptr = 0;
}
if(bf_rx0_start == 1)
{
byte_rx0_buffer[byte_rx0_ptr++] = rx_byte;
sprintf(buffer,"%X",rx_byte); //ASCII CONVERSION
dat=strtol(buffer,&pEnd,16);
// P1=(int)dat;
// sprintf(P1,"%s",dat);
delay_ms(2000);
k[byte_rx0_ptr++]=dat;
}
if ((byte_rx0_ptr == 14))
bf_rx0_start = 0;//end further rx until detect new STX
}
}
}
convert this value to hexadec value & xor it ie(3F^30^31^53^52^57=68), if i can do this calculation in program You fundamentally don't understand the difference between values and encodings. Two plus three is five whether you represent the two as "2", "two", or "X X". Addition operates on values, not representations. So to "convert to hexadecimal & xor it" makes no sense. You XOR values, not representations. Hexadecimal is a representation. To maintain a running XOR, just do something like int running_xor=0; at the top and then running_xor ^= rx_byte; each time you receive a byte. It will contain the correct value when you are finished. Set it to zero to reset it. Get hexadecimal completely out of your head. That is just how those values are being printed for your consumption. That has nothing to do with the internal logic of your program which deals only in values.
You would do well to separate out the data validation from the data reception, even to the extent that you don't do it in the interrupt handler; it is likely to be better to buffer the data in the ISR unchecked and defer the data validation to the main code thread or a task-thread if you are using an RTOS. You certainly don't want to be calling heavy-weight library functions such as sprintf() or strtol() in an ISR!
Either way, here is a function that would take a pointer to a received string, and its length (to avoid an unnecessary strlen() call since you already know how many characters were received), and returns true if the checksum validates, and false otherwise. It has no restriction on data length - that would be performed by the calling function.
If you know that your checksum hex digits will always be either upper or lower-case, you can simplify the decodeHexNibble() function.
#include <stdint.h>
#include <stdbool.h>
uint8_t decodeHexNibble() ;
uint8_t decodeHexByte( char* hexbyte ) ;
uint8_t decodeHexNibble( char hexdigit ) ;
bool checkData( char* data, int length )
{
int data_len = length - 2 ;
char* bcc_ptr = &data[data_len] ;
uint8_t rx_bcc_val = 0 ;
uint8_t actual_bcc_val = 0 ;
int i = 0 ;
// Convert <cc> string to integer
rx_bcc_val = decodeHexByte( bcc_ptr ) ;
// Calculate XOR of <data>
for( i = 0; i < data_len; i++ )
{
actual_bcc_val ^= data[i] ;
}
return actual_bcc_val == rx_bcc_val ;
}
uint8_t decodeHexNibble( char hexdigit )
{
uint8_t nibble ;
if( hexdigit >= '0' && hexdigit <= '9' )
{
nibble = hexdigit - '0' ;
}
else if( hexdigit >= 'a' && hexdigit <= 'f' )
{
nibble = hexdigit - 'a' + 10 ;
}
else if( hexdigit >= 'A' && hexdigit <= 'F' )
{
nibble = hexdigit - 'A' + 10 ;
}
else
{
// Do something 'sensible' with invalid digits
nibble = 0 ;
}
return nibble ;
}
uint8_t decodeHexByte( char* hexbyte )
{
uint8_t byte = hexbyte[0] << 4 ;
byte |= hexbyte[1] ;
return byte ;
}
Determine Position of Most Signifiacntly Set Bit in a Byte
I have a byte I am using to store bit flags. I need to compute the position of the most significant set bit in the byte.
Example Byte: 00101101 => 6 is the position of the most significant set bit
Compact Hex Mapping:
[0x00] => 0x00
[0x01] => 0x01
[0x02,0x03] => 0x02
[0x04,0x07] => 0x03
[0x08,0x0F] => 0x04
[0x10,0x1F] => 0x05
[0x20,0x3F] => 0x06
[0x40,0x7F] => 0x07
[0x80,0xFF] => 0x08
TestCase in C:
#include <stdio.h>
unsigned char check(unsigned char b) {
unsigned char c = 0x08;
unsigned char m = 0x80;
do {
if(m&b) { return c; }
else { c -= 0x01; }
} while(m>>=1);
return 0; //never reached
}
int main() {
unsigned char input[256] = {
0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,
0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17,0x18,0x19,0x1a,0x1b,0x1c,0x1d,0x1e,0x1f,
0x20,0x21,0x22,0x23,0x24,0x25,0x26,0x27,0x28,0x29,0x2a,0x2b,0x2c,0x2d,0x2e,0x2f,
0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x3b,0x3c,0x3d,0x3e,0x3f,
0x40,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,
0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x5b,0x5c,0x5d,0x5e,0x5f,
0x60,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f,
0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x7b,0x7c,0x7d,0x7e,0x7f,
0x80,0x81,0x82,0x83,0x84,0x85,0x86,0x87,0x88,0x89,0x8a,0x8b,0x8c,0x8d,0x8e,0x8f,
0x90,0x91,0x92,0x93,0x94,0x95,0x96,0x97,0x98,0x99,0x9a,0x9b,0x9c,0x9d,0x9e,0x9f,
0xa0,0xa1,0xa2,0xa3,0xa4,0xa5,0xa6,0xa7,0xa8,0xa9,0xaa,0xab,0xac,0xad,0xae,0xaf,
0xb0,0xb1,0xb2,0xb3,0xb4,0xb5,0xb6,0xb7,0xb8,0xb9,0xba,0xbb,0xbc,0xbd,0xbe,0xbf,
0xc0,0xc1,0xc2,0xc3,0xc4,0xc5,0xc6,0xc7,0xc8,0xc9,0xca,0xcb,0xcc,0xcd,0xce,0xcf,
0xd0,0xd1,0xd2,0xd3,0xd4,0xd5,0xd6,0xd7,0xd8,0xd9,0xda,0xdb,0xdc,0xdd,0xde,0xdf,
0xe0,0xe1,0xe2,0xe3,0xe4,0xe5,0xe6,0xe7,0xe8,0xe9,0xea,0xeb,0xec,0xed,0xee,0xef,
0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,0xfe,0xff };
unsigned char truth[256] = {
0x00,0x01,0x02,0x02,0x03,0x03,0x03,0x03,0x04,0x04,0x04,0x04,0x04,0x04,0x04,0x04,
0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,
0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,
0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,
0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,
0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,
0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,
0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08};
int i,r;
int f = 0;
for(i=0; i<256; ++i) {
r=check(input[i]);
if(r !=(truth[i])) {
printf("failed %d : 0x%x : %d\n",i,0x000000FF & ((int)input[i]),r);
f += 1;
}
}
if(!f) { printf("passed all\n"); }
else { printf("failed %d\n",f); }
return 0;
}
I would like to simplify my check() function to not involve looping (or branching preferably). Is there a bit twiddling hack or hashed lookup table solution to compute the position of the most significant set bit in a byte?
Your question is about an efficient way to compute log2 of a value. And because you seem to want a solution that is not limited to the C language I have been slightly lazy and tweaked some C# code I have.
You want to compute log2(x) + 1 and for x = 0 (where log2 is undefined) you define the result as 0 (e.g. you create a special case where log2(0) = -1).
static readonly Byte[] multiplyDeBruijnBitPosition = new Byte[] {
7, 2, 3, 4,
6, 1, 5, 0
};
public static Byte Log2Plus1(Byte value) {
if (value == 0)
return 0;
var roundedValue = value;
roundedValue |= (Byte) (roundedValue >> 1);
roundedValue |= (Byte) (roundedValue >> 2);
roundedValue |= (Byte) (roundedValue >> 4);
var log2 = multiplyDeBruijnBitPosition[((Byte) (roundedValue*0xE3)) >> 5];
return (Byte) (log2 + 1);
}
This bit twiddling hack is taken from Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup where you can see the equivalent C source code for 32 bit values. This code has been adapted to work on 8 bit values.
However, you may be able to use an operation that gives you the result using a very efficient built-in function (on many CPU's a single instruction like the Bit Scan Reverse is used). An answer to the question Bit twiddling: which bit is set? has some information about this. A quote from the answer provides one possible reason why there is low level support for solving this problem:
Things like this are the core of many O(1) algorithms such as kernel schedulers which need to find the first non-empty queue signified by an array of bits.
That was a fun little challenge. I don't know if this one is completely portable since I only have VC++ to test with, and I certainly can't say for sure if it's more efficient than other approaches. This version was coded with a loop but it can be unrolled without too much effort.
static unsigned char check(unsigned char b)
{
unsigned char r = 8;
unsigned char sub = 1;
unsigned char s = 7;
for (char i = 0; i < 8; i++)
{
sub = sub & ((( b & (1 << s)) >> s--) - 1);
r -= sub;
}
return r;
}
I'm sure everyone else has long since moved on to other topics but there was something in the back of my mind suggesting that there had to be a more efficient branch-less solution to this than just unrolling the loop in my other posted solution. A quick trip to my copy of Warren put me on the right track: Binary search.
Here's my solution based on that idea:
Pseudo-code:
// see if there's a bit set in the upper half
if ((b >> 4) != 0)
{
offset = 4;
b >>= 4;
}
else
offset = 0;
// see if there's a bit set in the upper half of what's left
if ((b & 0x0C) != 0)
{
offset += 2;
b >>= 2;
}
// see if there's a bit set in the upper half of what's left
if > ((b & 0x02) != 0)
{
offset++;
b >>= 1;
}
return b + offset;
Branch-less C++ implementation:
static unsigned char check(unsigned char b)
{
unsigned char adj = 4 & ((((unsigned char) - (b >> 4) >> 7) ^ 1) - 1);
unsigned char offset = adj;
b >>= adj;
adj = 2 & (((((unsigned char) - (b & 0x0C)) >> 7) ^ 1) - 1);
offset += adj;
b >>= adj;
adj = 1 & (((((unsigned char) - (b & 0x02)) >> 7) ^ 1) - 1);
return (b >> adj) + offset + adj;
}
Yes, I know that this is all academic :)
It is not possible in plain C. The best I would suggest is the following implementation of check. Despite quite "ugly" I think it runs faster than the ckeck version in the question.
int check(unsigned char b)
{
if(b&128) return 8;
if(b&64) return 7;
if(b&32) return 6;
if(b&16) return 5;
if(b&8) return 4;
if(b&4) return 3;
if(b&2) return 2;
if(b&1) return 1;
return 0;
}
Edit: I found a link to the actual code: http://www.hackersdelight.org/hdcodetxt/nlz.c.txt
The algorithm below is named nlz8 in that file. You can choose your favorite hack.
/*
From last comment of: http://stackoverflow.com/a/671826/315052
> Hacker's Delight explains how to correct for the error in 32-bit floats
> in 5-3 Counting Leading 0's. Here's their code, which uses an anonymous
> union to overlap asFloat and asInt: k = k & ~(k >> 1); asFloat =
> (float)k + 0.5f; n = 158 - (asInt >> 23); (and yes, this relies on
> implementation-defined behavior) - Derrick Coetzee Jan 3 '12 at 8:35
*/
unsigned char check (unsigned char b) {
union {
float asFloat;
int asInt;
} u;
unsigned k = b & ~(b >> 1);
u.asFloat = (float)k + 0.5f;
return 32 - (158 - (u.asInt >> 23));
}
Edit -- not exactly sure what the asker means by language independent, but below is the equivalent code in python.
import ctypes
class Anon(ctypes.Union):
_fields_ = [
("asFloat", ctypes.c_float),
("asInt", ctypes.c_int)
]
def check(b):
k = int(b) & ~(int(b) >> 1)
a = Anon(asFloat=(float(k) + float(0.5)))
return 32 - (158 - (a.asInt >> 23))