CODEWITHSUNDEEP
cbinarycharshiftbitwise-and
Astoach167
Astoach167

Reputation: 91

Why is the & 1 and + '0' is needed to display the binary number?

I got this code here to print the binary of a decimal, if I ran this function with argument as 3, it would print 0000 0011, which is right, I understand that >> will shift the binary to the right 7 to 0 to display the binary, but I do not understand the purpose of the code: & 1 and + 0, can someone tell me what are those for?

void gal_print(gal8 a)
{
    int i = 8;
    while (i--)
       // printf("%d", i);
        putchar((a >> i & 1) + '0');
}

Upvotes: 2

Views: 101

Answers (2)

0___________
0___________

Reputation: 67713

It actually does not have to ... (no &, no + '0')

void gal_print(unsigned char a)
{
    int i = CHAR_BIT;
    while (i--)
    {
        if(((unsigned char)(((a >> i) << (CHAR_BIT - 1))) >> (CHAR_BIT - 1)))
        {
            putchar('1');
        }
        else
        {
            putchar('0');
        }
    }

}

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 311048

This expression with the bitwise operator & (bitwise AND operator)

a >> i & 1

is used to extract the right most bit of the number. So the result value of the expression will be either 0 or 1.

For example

00000011 // 3
&
00000001 // 1
========
00000001 // 1

or

00000010 // 2
&
00000001 // 1
========
00000000 // 0

As there is used the function putchar then this integer value is required to be converted to character.

putchar((a >> i & 1) + '0');

That is '0' + 0 gives the character '0' and '0' + 1 gives the character '1'.

Upvotes: 3

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