What does (1U << X) do?

Question

I found this piece of code:

enum  {   IsDynamic = (1U << 0),  // ...   IsSharable = (1U << 1), // ...   IsStrong = (1U << 2)    // ... }; 

What does the (1U << X) do?

Answer 1

It sets bitmasks:

1U << 0 = 1 1U << 1 = 2 1U << 2 = 4 etc... 

What happens is 1U (unsigned value 1) is shifted to the left by x bits.

The code you posted is equivalent to:

enum  {       IsDynamic = 1U,  // binary: 00000000000000000000000000000001       IsSharable = 2U, // binary: 00000000000000000000000000000010       IsStrong = 4U    // binary: 00000000000000000000000000000100 } 

Answer 2

Bit shift. Instead of saying a = 1, b = 2, c = 4 they shift the bits. The idea is to pack many flags into one integer (or long).

This is actually a very clean approach.