十进制到二进制的转换方法

Question

I came across this method of conversion decimal numbers to binary:

int main(void){
    int i            = 0;
    unsigned int u_i = 0;
    int j            = 0;
    int b[16]        = {0}; //Assuming your integer size is 16bit

    printf("input number");
    scanf("%d",&i);

    u_i = (unsigned int)i;

    for(j=15;j>=0;j--) {
        b[j] = u_i & 0x1;
        u_i  = u_i >> 1;
    }

    for (j=0;j<=15;j++)
        printf("%d", b[j]);

    printf("\n");
    return 0;
}

I cannot understand the "0x1" part that is responsible for the conversion. Could anyone elaborate this? Thank you in advance.

Answer 1

What is happening is something called a mask. Basically what is going on is you have an unsigned integer of 16 bits. Lets say we have the following:

uint x = 43981

this is equivalent in hex to 0xABCD

And in binary 1010101111001101

So in order to convert this into an array, what the algorithm is doing is saying:

Shift the bits over one spot to the right.

1010101111001101 >> 1 becomes 0101010111100110

which is all of the numbers moved to the right and a 0 is placed on the left.

Then in order to just get the least significant bit, the following is done:

0101010111100110 
&
0000000000000001
=
0000000000000000 
= 
0

Another example would be

0101010111100111
&
0000000000000001
=
0000000000000001
=
1

Basically the & with just a 1 will make all of the other bits 0 no matter what, and then the one bit with a 1 will become the value of whatever bit is in the corresponding spot in the original number.

This will keep happening, and the original number will continue to be right shifted, until all of the numbers have been cycled through, creating your array of 0s and 1s.