为什么只在转换一个32位值时只使用移位操作的较低的5位?(UInt32)1 << 33 == 2)

Consider the following code:

考虑下面的代码:

UInt32 val = 1;
UInt32 shift31 = val << 31;                    // shift31  == 0x80000000
UInt32 shift32 = val << 32;                    // shift32  == 0x00000001
UInt32 shift33 = val << 33;                    // shift33  == 0x00000002
UInt32 shift33a = (UInt32)((UInt64)val << 33); // shift33a == 0x00000000
UInt3