CODEWITHSUNDEEP
cbit-manipulation
CHID
CHID

Reputation: 1643

rotation of unsigned char by n bits

I am trying to rotate an unsigned char by 'n' bits. But I am not getting the desired result. This is my code

void left_rotate(unsigned char a, int no){
        // no - number of times to rotate
        printf("%d\n", ((a << no) | (a >> (8-no))));
}

I am calling this function from mail as follows

unsigned char a = 'A';
left_rotate(a, 2);

I expected the following output 

//'A' = 65 = 01000001
// I am rotating in left direction by two times
// a << 2 = 00000100
// a >> 6 = 00000001
(00000100 | 00000001 = 00000101 = 5 in decimal)

But I got a different output

// The output in my screen = 100000101 = 261 in decimal

How did that 1 in MSB creep in? I am using an unsigned char as data type. So it shouldnt exceed 8 bits. Can someone please explain this?

Thanks

Chid

Upvotes: 6

Views: 1130

Answers (1)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727047

Since << promotes its arguments to unsigned int, you need to mask off the upper bits of the shift result:

printf("%d\n", (((a << no) & 0xFF) | (a >> (8-no))));

Demo on ideone (prints 5).

Upvotes: 10

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