Conversion of decimal to binary in C programming n store in arrays

[Zenoba]

Can someone give me an idea to convert decimal numbers into its binary equivalent and store each binary in arrays in C programming language??

[boxfish]

In C, all the number types are, somewhere deep in your computer, actually stored as a row of binary bits, so I wouldn't call a number "decimal" unless it's decimal digits were stored in a string. But aside from that, to get a bit at a specific position in an int, you can make a int that has only a bit in the place in question, compare it to the original int with the bitwise and operator (so that the result has all false bits unless both the ints have a bit in the same position), and then cast the result to a bool. Hope this helps.

[r035198x]

What ideas do YOU have for it yet?

[Zenoba]

What ideas do YOU have for it yet?

i have idea for only one decimal number
that is input a number then the program gives you its binary equivalent
the code is as follows:

int num,no_bits;

printf("Enter number");
scanf("%d",&num);
no_bits=count_bits(num);
printf("number of bits %d",no_bits);
bin_print(num,no_bits);

count_bits(int n) //function to count number of bits 
{
int i=0;
for (;n!=0;n=n>>1)
i++;
return(i);
}
bin_print(int x, int n_bits)
/* function to print decimal numbers in binary format */
{
int j;
printf("no. %d in binary \t",x);
for(j=n_bits-1; j>=0;j--)
printf("%i",(x>>j) & 01);
}

but that's only for one number
i have to do it for a series of numbers stored in an array and store the binary ones in another array

[weaknessforcats]

There is a simple algorithm for this.

Given the number 22 which has these bits:

16 8 .. 4 .. 2 .. 1
1.. 0 .. 1 .. 1 .. 0

To see if the bit in the 4 position is set, divide the number by 4 and take the result %2. That is:

22/4 = 5 and 5%2 is 1 --> the 4 bit is set
22/8 = 2 and 2%2 is 0 --> the 8 bit is not set

No need for fancy bit shifting.

[whodgson]

My book C++ In Plain English by Brian Overland has an interesting function for this:

void print_binary(short input_field)
{
    int i,bit_set;
    
    while (i<=16)
    { bit_set = ((Ox8000 & input_field) >0);
      printf("%d",bit_set);
      input_field = input_field  << 1;
      i++;}
}

I presume the equivalent can be written in C

[whodgson]

This is a function which emulates weaknessforcats ' algorithm

int convert(int n)
{int k=1;
 while(k<n)//find the most significant bit
  k*=2;
 if(k>n)//fix the overshoot
  k=k/2;

 while(k>0)
 {
 if(int(n/k)%2==0)cout<<"0 ";
 else cout<<"1 ";
 k=k/2;
 }
}