Strange result with bitshift operations

Question

I am trying to understand bishift operations better so I wrote myself a little program.

unsigned char a = 240;
a= (a << 3) >> 7;
printf("a: %u\n",a);

Now I would imagine that the result would be something like :

11110000 // 240
10000000 // << 3
00000001 // >> 7

So 1, but I get 15. I am confused... Any help is appreciated!

Answer 1

Your problem is that this statement : (a << 3) converts the input to an int . So at this point you have 240 * 2 ^ 3 = 1920

00000000000000000000011110000000

Then you are dividing the previous result by 2 ^ 7 = 128 so you have : 15

00000000000000000000000000001111

Which is exactly what you are getting as a result.

If you wanted to truncate bits you could have used :

printf("a: %u\n",a & 1); //get only last bit so you would have 1 as a result!
printf("a: %u\n",a & 255); //get all 8 bits

Hope this helped!

Answer 2

The expressions are evaluated as (unsigned) ints. (default int promotion). Casting(truncation) to a narrower type only happens just prior to the final assignment.