Dec to Bin

**hk_mp5kpdw** · 10-04-2004

Originally Posted by devour89

my only question left is can I seperate all the '0' and '1' into an array?

Array of characters, or array of integers 1s and 0s? One way perhaps (for a character array), use the bitset to_string() member function along with the string c_str() member function to get a c-style character string representation of the bitset that you can then strcpy into a temp array.

Code:

#include <cstring>
#include <bitset>
#include <iostream>
#include <string> // Not sure if needed in this instance, just in case

int main()
{
    std::bitset<8> bits(218);
    char array[9];  // Extra space to hold NULL copied during strcpy

    strcpy(array,bits.to_string().c_str());

    std::cout << array << std::endl;

    return 0;
}

Again, should output: 11011010

Edit: Forgot to mention that you can use the array indexing operator [] to access individual bits from the bitset in an array like manner which can be used in a loop to initialize an array. I.e. for the bitset in the above example we have:

bits[0] = 0
bits[1] = 1
bits[2] = 0
bits[3] = 1
bits[4] = 1
bits[5] = 0
bits[6] = 1
bits[7] = 1

Using that principal might be useful for you.

**elad** · 10-04-2004

Without using bitsets you could do this:

Code:

int input;
//get input
int x = input;
int array[10];
int i = 0;
while(x > 0)
{
  array[i++] = x % 2;
  x/2;
}

That will put the binary value of input in array in "backward" order where the power of 2 is the index of the element. For example:

input = 26 would come out as 01011 if printed in ascending order of index, and 11010 if printed in descending order of index---which could be readily accomplished.

**pablo615** · 10-04-2004

To get workable values from the char array, in a loop subtract the ascii value of '0' (0x30). '1' is 0x31...
Or, instead of coverting to a string, in a loop, use a bitmask like this...

Code:

...
        std::bitset<8> bits(218);
	int mask = 1;
	unsigned long bitArray[8] = {0};
	unsigned long bl = bits.to_ulong(); 
	for (int i = 7; i >=0 /*bad magic # :) */; i--, mask = mask << 1)
	{
		bitArray[i] = (bl & mask) >> (7 - i);
		printf("%d", bitArray[i]);
	}
	printf ("\n");
	return 0;

Or if you want to keep everything it bitset, create the bitArray as type bitset[8] and then get rid of the extra ulong bl. Just & the mask against bits but keep everything else the same. This will put the LSB in index 0.

PK

**pablo615** · 10-04-2004

Elad, x/2 needs to be x/=2...you've got yourself a nice inifinte loop there

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