FAQ: What does that final condition in the JPG signature check mean?

if (buffer[0] == 0xff && buffer[1] == 0xd8 && buffer[2] == 0xff && (buffer[3] & 0xf0) == 0xe0)

The & is the bitwise AND operator. It can be used as a “mask” to let us concentrate on only some bits in a number. When you AND two bits together, the result is 1 if both bits are set to 1, otherwise it’s 0.

Per the spec:

The fourth byte, meanwhile, is either 0xe0, 0xe1, 0xe2, 0xe3, 0xe4, 0xe5, 0xe6, 0xe7, 0xe8, 0xe9, 0xea, 0xeb, 0xec, 0xed, 0xee, or 0xef. Put another way, the fourth byte’s first four bits are 1110.

So, in other words, we want to check that the first 4 bits of the byte are 1110. We can do this by ‘masking’ away the final four bits. The value 0xf0 (11110000) works well for this since it has all first 4 bits and last 4 bits off. So the result of any number ANDed with it will have its last 4 bits off no matter what, so we can concentrate on the first 4 bits only.

Let’s look at an example. Say that buffer[3] is equal to 0xe6.

0xe6        11100110
0xf0        11110000
0xe6 & 0xf0 11100000 (that's 0xe0)  so that's a match

or, say that buffer[3] is 0xc7

0xc7        11000111
0xf0        11110000
0xc7 & 0xf0 11000000 (that's 0xc0) so that's not a match

Using the bitwise AND in our condition is a quick way to compare one number to a range of others in one step.

More information about bitwise in C: Bitwise Operations in C - wikipedia