CODEWITHSUNDEEP
cbit-manipulationstrtoull
Juhi
Juhi

Reputation: 1

Why first two digits are lost after using left-shift bitwise operator?

I am storing a hexadecimal value in a uint64_t variable. When I call strtoull() to convert the string to a hexadecimal value I get the correct result, but when I do a left-shift (<<) by 32 I am losing the first two digits.

The string I am testing is "398a59b412". I expect to get back 398a59b41200000000 but instead of this I get 8a59b41200000000.

#include <openssl/aes.h>
#include <openssl/modes.h>
#include <openssl/rand.h>
#include <openssl/hmac.h>
#include <openssl/buffer.h>
#include <stdio.h>
#include <inttypes.h>
#include <stdint.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
#define u8 unsigned char
#define INT_BITS 32 

unsigned long long int ecountGeneration()
{
   char *test_count = "398a59b412";
   unsigned long long x = 0x398a59b412 | 0x00;
   const char *test_bearer = "15";
    const char *test_direction = "1";
    unsigned long long result;
    int64_t result_count = 0;
     
     

  //errno = 0;
   result_count = strtoull(test_count, '\0', 16);
  uint64_t  result_bearer = strtoull(test_bearer, NULL, 16);
  uint64_t  result_direction = strtoull(test_direction, NULL, 16);
  result_count = result_count  | 0x0;
  uint64_t x1 = result_count<<32;
  printf("-->%" PRIx64 "\n" ,x1);
  uint64_t x2 = result_bearer<<27;
  uint64_t x3 = result_direction<<26;
  uint64_t x4 = x1 | x2 | x3;
  
  printf("----->%" PRIu64 "\n" , x1);


  return result;
}

Upvotes: 0

Views: 124

Answers (1)

Jiho Lee
Jiho Lee

Reputation: 987

Because 0x398a59b412 is 40 bits in size!

In hexadecimal, each digit can have a range of 0 thru 15, which requires 4 bits (aka a 'nibble'). 0x398a59b412 has 10 digits, so it is 40 bits size. Further, the << operator in C is not a 'cyclic shift', so the leftmost two nibbles are truncated (lost).

Upvotes: 1

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