converting variable to bits?

**draggy** · 07-12-2005

hi...

let's say i got a char a = 'A';
and an arry of bool b[8];

how to convert variable a into binary and store it into b[]?

Thanks

**Salem** · 07-12-2005

A for loop, and use of the >>(shift-right) and &(bitwise-and) operators

**draggy** · 07-12-2005

can you demostrate with some coding?

**mitakeet** · 07-12-2005

Code:

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

void printBits(unsigned long val){
    int i, end = sizeof(unsigned long) * 8;
    unsigned long tmp = (ULONG_MAX / 2) + 1;
    for (i=0; i<end; i++){
        if (tmp & val) putc('1', stdout);
        else putc('0', stdout);
        tmp >>= 1;
    }
}

int main(){
    int number = 0;
    char buf[BUFSIZ];

    while (1){
        printf("Please enter a number to convert to binary: ");
        fgets(buf, BUFSIZ, stdin);
        rewind(stdin);
        number = atoi(buf);
        printBits(number);
        printf("\n");
    }
    return 0;

}

**laserlight** · 07-12-2005

For example:
num & 1 is true if the 1st bit is 1.
Of course if it isnt true, then the 1st bit is 0.
num = num >> 1 would then bitwise shift num such that the 2nd bit is now the 1st bit.

**dwks** · 07-12-2005

Excellent code, mitakeet. Except I don't see BUFSIZ declared anywhere. Or is it part of one of the header files?

**draggy** · 07-12-2005

ok.. my code:

Code:

void main()
{
	unsigned char a = 1;
	unsigned char b[8];
	char mask = 1;
	int i;
	
	for(i = 8; i > 0; i--)
	{
		printf("%d", a & mask);
		b[i - 1] = a & mask;
		a >>= 1;
	}

	printf("\n");

	for(i = 0; i < 8; i++)
	{
		if(b[i] == 1)
		{
			printf("1");
		}
		else
		{
			printf("0");
		}
	}
}

ok... now how do we convert from bits back to the original variable value?

**mitakeet** · 07-12-2005

BUFSIZ is declared in <stdio.h> and is typically around 512 bytes.

If you want to put it back together, you |= with the appropriate bit mask.

**draggy** · 07-12-2005

coding example please....

**mitakeet** · 07-12-2005

int value = 0;
int mask1 = 1;
value |= mask1;//now first (lowest order) bit is set

int mask2 = 2
value |= mask2;//now first and second low order bits are set

int mask3 = 4;....

**draggy** · 07-12-2005

based on my program the bits is store on b[8];
i try:

Code:

void main()
{
	unsigned char a = 'a';
	unsigned char c = 0;
	unsigned char b[8];
	char mask = 1;
	int i;
	
	for(i = 8; i > 0; i--)
	{
		printf("%d", a & mask);
		b[i - 1] = a & mask;
		a >>= 1;
	}

	printf("\n");

	for(i = 0; i < 8; i++)
	{
		c |= b[i];
		c <<= 1;

		if(b[i] == 1)
		{
			printf("1");
		}
		else
		{
			printf("0");
		}		
	}

	printf("\n%c\n", c);
}

but not working......

**dwks** · 07-12-2005

What does it output?

**draggy** · 07-12-2005

output:

Code:

10000110
01100001
┬

Thanks

**dwks** · 07-12-2005

01100001 is the ASCII code for 'a'.

**draggy** · 07-12-2005

Originally Posted by dwks

01100001 is the ASCII code for 'a'.

yes... the binary ASCII

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