C array of binary values

Question

This is my c code that is within the function main:

int numbers[8] = {0,1,1,0,1,0,0,1}; 
int i = 7;
int value = 0;
while (i > -1){
   if (i == 7){
      if (numbers[i] == 1){
         value += 1; 
      }
   } else if (i == 6){
      if (numbers[i] == 1){
           value += 2;
      }
  } else if (i == 5){
      if (numbers[i] == 1){
            value += 4;
      }
  } else if (i == 4){
      if (numbers[i] == 1){
            value += 8;
      }
  } else if (i == 3){
      if (numbers[i] == 1){
             value += 16;
      }
  } else if (i == 2){
      if (numbers[i] == 1){
         value += 32;
      }
 } else if (i == 1){
      if (numbers[i] == 1){
           value += 64;
     }
 } else if (i == 0){
     if (numbers[i] == 1){
         value += 128;
     }
  }
  i--; 
}
printf("%d\n", value);

Is there a better way to display the decimal number that is within the array numbers? The array numbers represents the decimal value: 105, in binary.

There's a better way to do everything else in that program, but no, printf("%d") is a perfectly fine way to print the decimal value of a number. — Lee Daniel Crocker
– Lee Daniel Crocker, Commented Oct 31, 2017 at 23:53
I mean more so the way that I loop over the array and count the binary values. It seems like to many if statements. — Trevor Good
– Trevor Good, Commented Oct 31, 2017 at 23:54
Your instincts there are good...yes, it's too much repetition. Use a switch, or a loop. — Lee Daniel Crocker
– Lee Daniel Crocker, Commented Oct 31, 2017 at 23:55

Steve · Accepted Answer · 2017-10-31 23:54:51Z

2

int multiplier = 1;
int value = 0;
for (int i = 7; i >= 0; i--)
{
    value += (numbers[i] * multiplier);
    multiplier *= 2;
}

printf("%d\n", value);

answered Oct 31, 2017 at 23:54

Steve

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Vlad from Moscow · Accepted Answer · 2017-11-01 08:06:04Z

1

You can write a separate function that will convert an array of binary digits to a number. For example

#include <stdio.h>

unsigned long long int array_to_binary(const unsigned int a[], size_t n)
{
    const unsigned long long Base = 2;
    unsigned long long int binary = 0;

    for (size_t i = 0; i < n; i++)
    {
        binary = Base * binary + a[i];
    }

    return binary;
}

int main( void )
{
    unsigned int a[] = { 0, 1, 1, 0, 1, 0, 0, 1 };

    printf("array_to_binary( a, sizeof( a ) / sizeof( *a ) ) = %lld\n",
        array_to_binary(a, sizeof(a) / sizeof(*a)));

    return 0;
}

The program output is

array_to_binary( a, sizeof( a ) / sizeof( *a ) ) = 105

answered Nov 1, 2017 at 8:06

Vlad from Moscow

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