What does “<<” (double angle brackets) mean in C/C++ enum?

Question

enum ofp10_port_state {

    OFPPS10_STP_LISTEN  = 0 << 8, /* Not learning or relaying frames. */
    OFPPS10_STP_LEARN   = 1 << 8, /* Learning but not relaying frames. */
    OFPPS10_STP_FORWARD = 2 << 8, /* Learning and relaying frames. */
    OFPPS10_STP_BLOCK   = 3 << 8, /* Not part of spanning tree. */
    OFPPS10_STP_MASK    = 3 << 8  /* Bit mask for OFPPS10_STP_* values. */

};

Answer 1

Its a left bit shift operator. Meaning it shifts the bits left the indicated number of bits:

say that the value is:

0x0F or 00001111
0x0F << 4 = 0xF0 or 11110000

In microsoft c++ shifts right (>>) keep the sign (or the most significant digit, the one on the far left) depending on if the number is signed or unsigned

(assuming size of a byte):

signed integer (an int for example):
0x80 or 10000000
0x80 >> 7 = 11111111
0x10 or 00010000
0x10 >> 4 = 00000001
if its unsigned (a uint):
0x80 or 10000000
0x80 >> 7 = 00000001
0x10 or 00010000
0x10 >> 4 = 00000001

Answer 2

<< is a left bitshift operator.

If you have a bit pattern like 0010 (2 in decimal) and shift it to the left by 2 like so 0010<<2 , you get 1000 (8 in decimal).

An enum is simply an integer that is large enough to hold at least an int . Thus we can directly assign int values like 0, 1, etc. to it.

In this case, we are assigning things like 1 << 8 to it (which yields 100000000 or 256 in decimal).