binary numbers

how do you convert a binary number to its decimal equivalent? if someone could explain this to me in English that would be helpful. I'm not quite sure what a binary number is or how to go about creating a program that does the above. any feedback would be appreciated.

http://www.permadi.com/tutorial/numBinToDec/

Binary numbers do take a bit of getting used to, but it's not too bad once one figures out what's going on. This is not scientific by any means.

Start with 1 byte or 8 bits.

1 means a bit is set (on)
0 means a bit is not set (off)

bits 0-7

bit 0 = 1..... 2^0
bit 1 = 2..... 2^1
bit 2 = 4..... 2^2
bit 3 = 8..... 2^3
bit 4 = 16.... 2^4
bit 5 = 32.... 2^5
bit 6 = 64.... 2^6
bit 7 = 128.. 2^7

A bit pattern is read from right to left, so you must remember to fill in all missing left side bits with zeros (as you become familiar with the concept, you won't need to do that.)

1101 read right to left is

bit 0 set = 1
bit 1 not set = 0
bit 2 set = 1
bit 3 set = 1

the full eight bits

00001101b

look back at the bits 0-7 and add the set bits together. You get 13 dec. The 7th bit is a sign bit.

As far as a proggie, that isn't too hard either provided that you keep in mind my example is very basic.

Code:

---

#include <stdio.h>
#include <string.h>
 
int main(void)
{
  int oneByte[8] = { 1, 2, 4, 8, 16, 32, 64, 128 };
               
  char binary[15]; /* 8 bits + '\0' + overflow margin */
  char buff[9];
  char *SB; /* sign bit */
  int i, j, decimal;
       
  printf("Enter a binary string up to (8 bits): ");
  gets(binary);
       
  if(strlen(binary) > 8)
  {
    printf("\n\nInput string greater than eight bits");
  return -1;
  }
       
  if(binary[0] == '\0')
  {
    printf("\n\nThere is nothing to do.");       
    return 0; /* not really an error */
  }
       
  j=0;
       
  /* reverse string */
       
  for(i=strlen(binary)-1; i >= 0; i--)
    buff[j++] = binary[i];
               
  /* terminate the reversed string */
       
  buff[j] = '\0';
       
  decimal = 0;
       
  for(i=0; buff[i] != '\0'; i++)
  {
    /* use the switch to find out if the data
is good */
               
  switch(buff[i])
  {
    case '0': break;
    case '1': decimal += oneByte[i]; break;
    case '\0': break;
    default :
    printf("\n\nInvalid entry => %c <= detected in input", buff[i]);
    return -1;
    break;
  };
                 
  }
       
  if(strlen(buff) == 8 && buff[7] == '1')
    SB = "(unsigned)";
  else
    SB = "(signed)";
       
  printf("\n\n%sb is %d dec %s", binary, decimal, SB);       
       
  return 0;
}

---

I *think* that will do... technically correct or not there ya go. Good luck.

Any given number can be viewed as a binary, octal, decimal, hexidecimal - whatever. These are just ways of looking at numbers. Your computer can't actually store a number any other way than binary. When the computer was invented, there was no way to feasably store 10 states (decimal) on a single "switch", so bi-state (binary) switches were chosen. Since, like a light bulb, only two states can be represented with a binary switch, the binary number system evolved. If you have two light bulbs, you can store the value 0 or 1. If you have two, a number as high as 3 can be stored. So on. Binary as it turns out is in many respects superior to any other number system.

you've helped me alot! my c class will be over as of thursday...so i wont be posting anymore...:(

but thank you all very much for helping me and puting up with my questions...:cool: