Evaluate 16-bit signed value after left shift by 2

Question

I am expecting zero value of 16-bit signed value 0x8000 after left shift by 2.

Following is my code

int main()
{
 register signed short val1 = 0x8000 ;
 signed short val2 ;
 val2 = ( ( val1 > 2 ) ; /* val2 should be 0. */
 printf( "nval2 is :%xn", val2 ) ; /* But Val2 is ffff8000 */
 return 0 ;
}

$ gcc myfile.c
$ ./a.out

I have verified the same behavior with 32-bit compilers and expected (val2=0) with 16-bit compiler. I can correct this by shifting 18 in place of 2 as a workaround. I want some verify my understanding for this point and another way to get expected result (if any).

Answer 1

signed short is 2 bytes(16 bits) long and it ranges from -32768 to +32767.
When you store 0x8000 in val1 ( SIGNED SHORT), it is actually considered a s a negative number because the sign bit is set high.
Binary representation - 1000 0000 0000 0000 for 0x8000.

You can verify it by printing the value of val1 as "%d" immediately after storing 0x8000 in it. It will print -32768.
So in 0xffff8000, FFFF indicates that it is a negaive number and 8000 is the value.
Similarly if you right shift this number by 2, the next 2 bits will be set high because of the signed bit copy mechanism. So val1 >>2 will give 0xFFFFe000 where FFFF signifies it is a negative number.
And 0x8000 -> 1000 0000 0000 0000 becomes 0xE000 -> 1110 0000 0000 0000 .

For more info you can follow this link
http://stackoverflow.com/questions/1857928/right-shifting-negative-numbers-in-c

Hope this helps.

Answer 2

Read this article which I wrote long back
http://tech.queryhome.com/45054/bitwise-shift-and-negative-numbers-in-c-c

When we right-shift then the filling is done as 0 if most significant bit is 0, one if most significant bit is one. Here the most significant bit is one so filling is done as one.