bit shift in c language

Question

I tried:

#include<stdio.h>

main()
{
    int a=10000;
    unsigned char cp1=0,cp2=0,cp3=0,cp4=0;

    cp1 = (a & 0xff000000) >> 24;
    cp2 = (a & 0x00ff0000) >> 16;
    cp3 = (a & 0x0000ff00) >>  8;
    cp4 = (a & 0x000000ff)      ;

    printf("%d %d %d %d\n",cp1,cp2,cp3,cp4);
}

My output is:

0 0 39 16

I found (39<<8) + 16=10000.

I could not understand cp3=(a & 0x0000ff00)>>8; ==39 how it is working?

I know 0xff=255, I want to know how the (&) operation and 0xff are working together and taking particular bits.

Can you teach me how it is working?

@OliCharlesworth; 7.21.6 P(9): If a conversion specification is invalid, the behavior is undefined.282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined. — haccks
– haccks, Commented Jun 20, 2014 at 5:50
@haccks: If all values of an unsigned char can be represented by a signed int (which is usually the case), the arguments will be promoted to signed int which matches the conversion specifiers. — Nisse Engström
– Nisse Engström, Commented Jun 20, 2014 at 6:07

Inder Kumar Rathore · Accepted Answer · 2014-06-20 06:09:54Z

2

             a  = 0010 0111 0001 0000
        0xff00  = 1111 1111 0000 0000
   (a & 0xff00) = 0010 0111 0000 0000
(a & 0xff00)>>8 = 0000 0000 0010 0111  //shift the bits of above ANDing 8 times to right
0000 0000 0010 0111 = 39 in decimal

edited Jun 20, 2014 at 6:09

answered Jun 20, 2014 at 5:59

Inder Kumar Rathore

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2 Comments

subramaniam Over a year ago

okay friend thanks for your help. you said clear cut answer.thanks again :D

Inder Kumar Rathore Over a year ago

My pleasure. Btw welcome to our community I hope you've read FAQ, it's important to read this and enjoyed the Q&A on SO. Cheers

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