I'm trying to figure out the purpose of this piece of code, from the Tiled utility's map format documentation.
const int gid = data[i] |
data[i + 1] << 8 |
data[i + 2] << 16 |
data[i + 3] << 24;
It looks like there is some "or-ing" and shifting of bits, but I have no clue what the aim of this is, in the context of using data from the tiled program.
As you noted, the << operator shifts bits to the left by the given number.
This block takes the data[] array, which has four (presumably one byte) elements, and "encodes" those four values into one integer.
Example Time!
data[0] = 0x3A; // 0x3A = 58 = 0011 1010 in binary
data[1] = 0x48; // 0x48 = 72 = 0100 1000 in binary
data[2] = 0xD2; // 0xD2 = 210 = 1101 0010 in binary
data[3] = 0x08; // 0x08 = 8 = 0000 1000 in binary
int tmp0 = data[0]; // 00 00 00 3A = 0000 0000 0000 0000 0000 0000 0011 1010
int tmp1 = data[1] << 8; // 00 00 48 00 = 0000 0000 0000 0000 0100 1000 0000 0000
int tmp2 = data[2] << 16; // 00 D2 00 00 = 0000 0000 1101 0010 0000 0000 0000 0000
int tmp3 = data[3] << 24; // 08 00 00 00 = 0000 1000 0000 0000 0000 0000 0000 0000
// "or-ing" these together will set each bit to 1 if any of the bits are 1
int gid = tmp1 | // 00 00 00 3A = 0000 0000 0000 0000 0000 0000 0011 1010
tmp2 | // 00 00 48 00 = 0000 0000 0000 0000 0100 1000 0000 0000
tmp3 | // 00 D2 00 00 = 0000 0000 1101 0010 0000 0000 0000 0000
tmp4; // 08 00 00 00 = 0000 1000 0000 0000 0000 0000 0000 0000
gid == 147998778;// 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
Now, you've just encoded four one-byte values into a single four-byte integer.
If you're (rightfully) wondering, why would anyone want to go through all that effort when you can just use byte and store the four single-byte pieces of data directly into four bytes, then you should check out this question:
int, short, byte performance in back-to-back for-loops
Bonus Example!
To get your encoded values back, we use the "and" operator along with the right-shift >>:
int gid = 147998778; // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
// "and-ing" will set each bit to 1 if BOTH bits are 1
int tmp0 = gid & // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
0x000000FF; // 00 00 00 FF = 0000 0000 0000 0000 0000 0000 1111 1111
int data0 = tmp0; // 00 00 00 3A = 0000 0000 0000 0000 0000 0000 0011 1010
int tmp1 = gid & // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
0x0000FF00; // 00 00 FF 00 = 0000 0000 0000 0000 1111 1111 0000 0000
tmp1; //value of tmp1 00 00 48 00 = 0000 0000 0000 0000 0100 1000 0000 0000
int data1 = tmp1 >> 8; // 00 00 00 48 = 0000 0000 0000 0000 0000 0000 0100 1000
int tmp2 = gid & // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
0x00FF0000; // 00 FF 00 00 = 0000 0000 1111 1111 0000 0000 0000 0000
tmp2; //value of tmp2 00 D2 00 00 = 0000 0000 1101 0010 0000 0000 0000 0000
int data2 = tmp2 >> 16; // 00 00 00 D2 = 0000 0000 0000 0000 0000 0000 1101 0010
int tmp3 = gid & // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
0xFF000000; // FF 00 00 00 = 1111 1111 0000 0000 0000 0000 0000 0000
tmp3; //value of tmp3 08 00 00 00 = 0000 1000 0000 0000 0000 0000 0000 0000
int data3 = tmp3 >> 24; // 00 00 00 08 = 0000 0000 0000 0000 0000 0000 0000 1000
The last "and-ing" for tmp3 isn't needed, since the bits that "fall off" when shifting are just lost and the bits coming in are zero. So:
gid; // 08 D2 48 3A = 0000 1000 1101 0010 0100 1000 0011 1010
int data3 = gid >> 24; // 00 00 00 08 = 0000 0000 0000 0000 0000 0000 0000 1000
but I wanted to provide a complete example.