Contents
Once I was roaming the Visual C++ forum (again), I had to face the fact that bitwise operations, and binary in general, are rarely in beginner's common sense. After having pained my fingers to write a very long answer to that innocent person, it became obvious that I had to share this obfuscated knowledge with the community through this article.
This is obviously a beginners article, but if you want to get deeper knowledge about the C/C++ bitwise operators cover, you can read the very complete article, An introduction to bitwise operators by PJ Arends. You can also go into a very complex (but effective) analysis of bit hacking with the article, Bit Twiddling Hacks By Sean Eron Anderson.
I'm going to present in the most complete way that I can about what we can do with bitwise operators, flags and all that binary stuff.
One of the places where we most find the use for this is certainly when a library provides a set of enumerations and when functions use DWORD
as a flags container. Let's take for that article the example of an enum
which defines some styles:
enum {
STYLE1 = 1,
STYLE2 = 2,
STYLE3 = 4,
STYLE4 = 8,
STYLE5 = 16,
STYLE6 = 32,
STYLE7 = 64,
STYLE8 = 128
};
 OR 
enum {
STYLE1 = 0x1,
STYLE2 = 0x2,
STYLE3 = 0x4,
STYLE4 = 0x8,
STYLE5 = 0x10,
STYLE6 = 0x20,
STYLE7 = 0x40,
STYLE8 = 0x80
};

As we can see, these constants are all powers of 2. To understand why these constants are chosen, we must go in binary representation:
1 > 0b 00000000 00000000 00000000 00000001
2 > 0b 00000000 00000000 00000000 00000010
4 > 0b 00000000 00000000 00000000 00000100
8 > 0b 00000000 00000000 00000000 00001000
16 > 0b 00000000 00000000 00000000 00010000
32 > 0b 00000000 00000000 00000000 00100000
64 > 0b 00000000 00000000 00000000 01000000
128 > 0b 00000000 00000000 00000000 10000000
Notice that for all of these values, only one bit
is set at a time, all the others are equal to 0
. You now can see a high level of interest appearing in this, which is that each bit is used as a flag for a functionality (here, each bit represents a style). We can now imagine a way to mix the flags together in one variable, to avoid using as many booleans as we need flags. Consider the following example :
0b 00000000 00000000 00000000 00100101
Flags of Style1, Style3 and Style6 are set
We face a problem now. C++ doesn't handle binary directly. We have to use bitwise operators
instead. There are 3 atomic bitwise operators to know, presented by ascending order of priority : OR (
), AND (&
) and NOT (~
). Here are their behaviors:
x y  x y & x ~
  
0 0 0 0 0 0 0 1
0 1 1 0 1 0 1 0
1 0 1 1 0 0
1 1 1 1 1 1
Knowing this, we can use these operators to build some mixes such as the ones presented above.
In the case of the instruction STYLE1  STYLE3  STYLE6
, we OR
the constants like this:
0b 00000000 00000000 00000000 00000001 < STYLE1
0b 00000000 00000000 00000000 00000100 < STYLE3
0b 00000000 00000000 00000000 00100000 < STYLE6

0b 00000000 00000000 00000000 00100101 < STYLE1  STYLE3  STYLE6
We can see that the bitwise OR
operator is pretty much like the addition (+) operator. However, you have to be very careful if you wished to use + instead or . The reason is simple: adding 1 + 1
resolves into 0
(plus one carry over 1
). You won't see any problem if all the constants are strictly different though, but I won't go deeper as it is a bad practice to use other than bitwise operators when processing binary operations.
Commonly, such mixes stick to the DWORD
type. However, this is not a must as we can do this with any integer types (char
, short
, int
, long
...). A DWORD
is an unsigned 32 bits integer (like those used in the binary representations of this article). Let's imagine the case when we have such a DWORD
constructed in a function, and passed to another in parameter. How can the called function know which bits are set, and which are not? Easy... Follow me!
Say we want to know if the bit of STYLE8
is set in the DWORD
passed in parameter. We must have a mask which we will call the AND
parameter. In practice, the mask is the same as the constant we are about to test, so there's no additional code to create such a mask:
DWORD parameter > 0b 00000000 00000000 00000000 00100101
STYLE8 mask > 0b 00000000 00000000 00000000 10000000

Bitwise AND > 0b 00000000 00000000 00000000 00000000 < 0x00000000
DWORD parameter > 0b 00000000 00000000 00000000 10100101
STYLE8 mask > 0b 00000000 00000000 00000000 10000000

Bitwise AND > 0b 00000000 00000000 00000000 10000000 < STYLE8
If the AND
operation returns 0
, then, the bits were not set, otherwise, you get the mask you applied.
OK, now, in practice, here is how you'll often see it:
void SetStyles(DWORD dwStyles) {
if ((STYLE1 & dwStyles) == STYLE1) {
//Apply style 1
}
else if ((STYLE2 & dwStyles) == STYLE2) {
//Apply style 2
}
else if ((STYLE3 & dwStyles) == STYLE3) {
//Apply style 3
}
//etc...
}
I didn't present the third operator NOT (~
) yet. It is generally used when you have a set of bits, in which some are set and some aren't, and where you'd like to remove one of them. The sample of code below exposes how this can be done:
void RemoveStyle5 (DWORD& dwStyles) {
if ((STYLE5 & dwStyles) == STYLE5) {
dwStyles = dwStyles & ~STYLE5;
}
}
I didn't mention the XOR ( ^
) operator yet. The reason why it comes last is for the only reason that this operator is not atomic; that means that we can reproduce its behavior with the other operators presented already:
#define XOR(a,b) (((a) & ~(b))  ((b) & ~(a)))
Anyway, this operator can be used to easily switch one bit :
void SwitchStyle5 (DWORD& dwStyles) {
dwStyles = dwStyles ^ STYLE5;
}
Now, to perfect your sharpen skills on bits handling, there are few other operators you have to know to be unbeatable: the shift operators. Those can be recognized like this : <<
(left shifting, follow the arrows), >>
(guess what, this is a right shifting).
What a shifting operator is moving the bits of n positions to the right or to the left. The "space" made by the movement is padded with zeros, and the bits pushed over the limit of the memory area are lost.
Let's take an example:
BYTE dwStyle = STYLE1  STYLE3  STYLE6; // [00100101]000
<
dwStyle = dwStyle << 3; // <s>001</s>[00101000]
BYTE dwStyle = STYLE1  STYLE3  STYLE6; // 000[00100101]
>
dwStyle = dwStyle >> 3; // [00000100]<s>101</s>
"But" I hear you say, "what is that for" ?". Well, there are tons of applications which I don't particularly have in mind right now, but trust me, when you need it, be happy it's there.
Anyway, two funny uses of such operators I can think of is the division and the multiplication of an integer by 2. Check this:
char i = 127; // 0b01111111 ( = 127)
i = i >> 1; // 0b00111111 ( = 63)
i = i << 1; // 0b01111110 ( = 126)
Shifting one bit right, then one bit left gives a different result from the beginning. That's because as we just saw, the "overflown" bits are lost, and the newly inserted ones are set to 0. But let's look closer. This is working anyway because we are working on integers. That is, dividing an odd number will not give a float (we don't care about the remaining 0.5).
Here, 127 / 2
should be 63.5
, so it gives 63
, due to the truncation.
63 * 2 = 126
, isn't it what we want ?! :)
Now we've seen all the useful operators, just know that all of them have their own assignment operator.
dwStyle &= 0x2 > dwStyle = dwStyle & 0x2
dwStyle = 0x2 > dwStyle = dwStyle  0x2
dwStyle ^= 0x2 > dwStyle = dwStyle ^ 0x2
dwStyle <<= 0x2 > dwStyle = dwStyle << 0x2
dwStyle >>= 0x2 > dwStyle = dwStyle >> 0x2
Thanks to Randor in the VC++ forum, here is a nice example of what is possible to do with such operators (Credit for the discovery of this little neat trick below goes to Sean Anderson at stanford university).
"How is it possible to reverse the bits of an integer, say from 11010001 to 10001011 ?"
BYTE b = 139; // 0b10001011
b = ((b * 0x0802LU & 0x22110LU)  (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16;
"Wowwww, how is this working man ?!"
DWORD iTmp = 0;
BYTE b = 139; // 00000000 00000000 00000000 10001011
DWORD c = (b * 0x0802LU); // 00000000 00000100 01011001 00010110
c &= 0x22110LU; // 00000000 00000000 00000001 00010000
DWORD d = (b * 0x8020LU); // 00000000 01000101 10010001 01100000
d &= 0x88440LU; // 00000000 00000000 10000000 01000000
iTmp = (c  d); // 00000000 00000000 10000001 01010000
iTmp = iTmp * 0x10101LU; // 10000001 11010001 11010001 01010000
iTmp >>= 16; // 00000000 00000000 10000001 11010001
b = iTmp; // 00000000 00000000 00000000 11010001
Well, here we are then. If you came here to understand these binary operations, I hope that I helped you in your way. If you came to see what was going on here, then don't hesitate to point my mistakes if any, so that I can fix them.
The following links are for complimentary knowledge, if you like to go deeper in bitwise handling: