Why use only the lower five bits of the shift operand when shifting a 32-bit value? (e.g. (UInt32)1 << 33 == 2)

Question

Consider the following code:

UInt32 val = 1;
UInt32 shift31 = val << 31;                    // shift31  == 0x80000000
UInt32 shift32 = val << 32;         


        

Answer 1

Unit32 overflows at 32 bits, that is defined in the spec. What are you expecting?

The CLR does not define a left shift with overflow detection operator(1). If you need that kind of facility you need to check for yourself.

(1) The C# compiler might cast it to long, but I am not sure.