Convert an Array of Ints Representing a Binary Number to a Decimal Number

Question

I have an array of ints that represent a binary number. I need to convert it to its decimal integer equivalent. Can someone please help me?

int octet[8] = {0, 1, 0, 0, 1, 0, 1, 0};

I want to return the decimal equivalent which would be 74 in this case.

Answer 1

This is an example of how you could do something like that:

#include <stdio.h>                                                                                                                                      
                                                                            
int main()                                                                      
{                                                                               
                                                                            
    int octet[] = {0, 1, 0, 0, 1, 0, 1, 0};                                    
    int *p = octet;                                                             
    int len = sizeof(octet) / sizeof(octet[0]); // 8                                                                
    unsigned result = 0;                                                        
                                                                                
    while(len--) result = (result<<1) | *p++;                                   
    /* 74 */                                                                     
    printf("%u\n", result);                                                     
                                                                                
}                   

Answer 2

2^0 x the last element + 2^1 x the second-to-last element + 2^2 x the third-to-last element +...

Answer 3

#include <stdio.h>

int main()                                                                      
{
  // Make array constant and let the compiler tell us how many elements are contained.
  // Doing it this way allows "octet" to grow up to the number of bits in an integer elements.  
  // It won't be an "octet" anymore though,...
  const int octet[] = {0, 1, 0, 0, 1, 0, 1, 0};         
  const int NumArrayElements = sizeof(octet) / sizeof(*octet);

  // Assign hi bit to save a loop iteration.
  int value = octet[0];
  
  // Run remaining bits, shifting up and or-ing in new values.
  for(int i=1; i<NumArrayElements; i++)
  {
    value <<= 1;
    value |= octet[i];
  }
  
  printf("%d\n", value);
}

This is a simple thing to do, but having the compiler handle some of the sizing issues and adding comments makes it much easier to maintain over time.