Trying to Rightshift 32 Bits

Question

why does [code]8103936 >> 32[/code] return 8103936? I need it to return 0 like it should. How do I fix

Alan · Answer

According to the C# spec the reason for this state of affairs is that for a 32 bit number only the 5 low order bits are significant in determining the number of bits to shift. When the result is 0 the number itself is returned 32 is of course 100000 in binary and so 0 bits are shifted and therefore the unchanged number is returned. For a 64 bit number the 6 low order bits are significant and we can therefore shift up to 63 bits before we start again at 0 which is why Mike's approach works.

Mike Gold · Answer

It appears that shifting it 31 bits returns 0 for 32 bit numbers. After that you are overriding the register. If you use long (64 bit) values you will get the desired results static void Main ( string [] args ) { long val1 = 8103936; int shift = 32; long val = val1 >> shift ; Console . WriteLine ( The value of {0} right shifted {1} is {2}.  val1  shift  val ); Console . ReadLine (); }