I came across a part of code that I cannot understand.
for (unsigned int i = (x & 0b1); i < x; i+= 2)
{
// body
}
Here, x is from 0 to 5.
What is meant by 0b1? and what would be the answers for eg: (0 & 0b1), (4 & 0b1) etc?
0b... is a binary number, just like 0x... is hex and 0... is octal.
Thus 0b1 is same as 1.
1b0 is illegal, the first digit in those must always be 0.
As previous answers said, it is the binary representation of the integer number 1, but they don't seem to have fully answered your question. This has a lot of layers so I'll briefly explain each.
In this context, the ampersand is working as a bitwise AND operator. i & 0b1 is (sometimes) a faster way of checking if an integer is even as opposed to i % 2 == 0.
Say you have int x = 5 and you'd like to check if it's even using bitwise AND.
In binary, 5 would be represented as 0101. That final 1 actually represents the number 1, and in binary integers it's only present in odd numbers. Let's apply the bitwise AND operator to 5 and 1;
0101
0001
&----
0001
The operator is checking each column, and if both rows are 1, that column of the result will be 1 – otherwise, it will be 0. So, the result (converted back to base10) is 1. Now let's try with an even number. 4 = 0100.
0100
0001
&----
0000
The result is now equal to 0. These rules apply to every single integer no matter its size.
The higher-level layer here is that in C, there is no boolean datatype, so booleans are represented as integers of either 0 (false) or any other value (true). This allows for some tricky shorthand, so the conditional if(x & 0b1) will only run if x is odd, because odd & 0b1 will always equal 1 (true), but even & 0b1 will always equal 0 (false).
I am a complete beginner at bitwise operations (and not very experienced at C either) and I bumped into the expression:
x |= (1<<y)
At first I thought it meant "x equals x or y shifted left by on bit", but then I realized that would be:
x |= (y<<1)
Lastly I thought it meant "x equals x or 1 shifted left by y bits", but I don't understand where that 1 is in an 8-bit register, does it mean 00000001? so that:
a = 2
b = 1<<a // so b=00000010
Could someone tell me the correct meaning of this statement. Also, if anyone has a good link explaining bitwise syntax I'd be grateful.
Thanks.
x |= ...
is shorthand for
x = x | ...
It assigns the value of x | ... to x.
1 << y
is 1 left-shifted by y. E.g.
00000001 << 1 -> 00000010
So,
x |= (1 << y)
is OR x with 1 left shifted by y (and assign the result to x).
In other words, it sets the y'th bit of x to 1.
x = 01010101
x |= (1 << 1) -> 01010111 (it set the 2nd bit to 1)
The first statement means left shift the binary representation of 1 (0b0000001) by y bits. Then OR the value with X.
The assumption is correct for the second statement.
The third statement will yield 4 (0b0000000100).
In terms of bit operation semantics the C standard defines all bit operations to represented such that binary numbers are read right to left with ascending values of powers of 2. You do not need to worry about endianess or two complements etc, the compiler will handle that for you. So (0b00100) = 4, (0b000010) = 2, (0b00001) = 1, and so on.
How would i go about accessing the individual bits inside a c++ type, char or any c++ other type for example.
If you want access bit N:
Get: (INPUT >> N) & 1;
Set: INPUT |= 1 << N;
Unset: INPUT &= ~(1 << N);
Toggle: INPUT ^= 1 << N;
You would use the binary operators | (or), & (and) and ^ (xor) to set them. To set the third bit of variable a, you would type, for instance:
a = a | 0x4
// c++ 14
a = a | 0b0100
Note that 4’s binary representation is 0100
That is very easy
Lets say you need to access individual bits of an integer
Create a mask like this
int mask =1;
now, anding your numberwith this mask gives the value set at the zeroth bit
in order to access the bit set at ith position (indexes start from zero) , just and with (mask<
If you want to look at the nth bit in a number you can use: number&(1<<n).
Essentially the the (1<<n) which is basically 2^n(because you shift the 1 bit in ...0001 n times, each left shift means multiply by 2) creates a number which happens to be 0 everywhere but 1 at the nth position(this is how math works).
You then & that with number. This returns a number which is either 0 everywhere or a number that has a 1 somewhere(essentially an integer which is either 0 or not).
Example:
2nd bit in in 4, 4&(1<<2)
0100
& 0010
____
0000 = 0
Therefore the 2nd bit in 4 is a 0
It will also work with chars because they are also numbers in C,C++
Hello is have a question for a school assignment i need to :
Read a round number, and with the internal binaire code with bit 0 on the right and bit 7 on the left.
Now i need to change:
bit 0 with bit 7
bit 1 with bit 6
bit 2 with bit 5
bit 3 with bit 4
by example :
if i use hex F703 becomes F7C0
because 03 = 0000 0011 and C0 = 1100 0000
(only the right byte (8 bits) need to be switched.
The lession was about bitmanipulation but i can't find a way to make it correct for al the 16 hexnumbers.
I`am puzzling for a wile now,
i am thinking for using a array for this problem or can someone say that i can be done with only bitwise ^,&,~,<<,>>, opertors ???
Study the following two functions:
bool GetBit(int value, int bit_position)
{
return value & (1 << bit_position);
}
void SetBit(int& value, int bit_position, bool new_bit_value)
{
if (new_bit_value)
value |= (1 << bit_position);
else
value &= ~(1 << bit_position);
}
So now we can read and write arbitrary bits just like an array.
1 << N
gives you:
000...0001000...000
Where the 1 is in the Nth position.
So
1 << 0 == 0000...0000001
1 << 1 == 0000...0000010
1 << 2 == 0000...0000100
1 << 3 == 0000...0001000
...
and so on.
Now what happens if I BINARY AND one of the above numbers with some other number Y?
X = 1 << N
Z = X & Y
What is Z going to look like? Well every bit apart from the Nth is definately going to be 0 isnt it? because those bits are 0 in X.
What will the Nth bit of Z be? It depends on the value of the Nth bit of Y doesn't it? So under what circumstances is Z zero? Precisely when the Nth bit of Y is 0. So by converting Z to a bool we can seperate out the value of the Nth bit of Y. Take another look at the GetBit function above, this is exactly what it is doing.
Now thats reading bits, how do we set a bit? Well if we want to set a bit on we can use BINARY OR with one of the (1 << N) numbers from above:
X = 1 << N
Z = Y | X
What is Z going to be here? Well every bit is going to be the same as Y except the Nth right? And the Nth bit is always going to be 1. So we have set the Nth bit on.
What about setting a bit to zero? What we want to do is take a number like 11111011111 where just the Nth bit is off and then use BINARY AND. To get such a number we just use BINARY NOT:
X = 1 << N // 000010000
W = ~X // 111101111
Z = W & Y
So all the bits in Z apart from the Nth will be copies of Y. The Nth will always be off. So we have effectively set the Nth bit to 0.
Using the above two techniques is how we have implemented SetBit.
So now we can read and write arbitrary bits. Now we can reverse the bits of the number just like it was an array:
int ReverseBits(int input)
{
int output = 0;
for (int i = 0; i < N; i++)
{
bool bit = GetBit(input, i); // read ith bit
SetBit(output, N-i-1, bit); // write (N-i-1)th bit
}
return output;
}
Please make sure you understand all this. Once you have understood this all, please close the page and implement and test them without looking at it.
If you enjoyed this than try some of these:
http://graphics.stanford.edu/~seander/bithacks.html
And/or get this book:
http://www.amazon.com/exec/obidos/ASIN/0201914654/qid%3D1033395248/sr%3D11-1/ref%3Dsr_11_1/104-7035682-9311161
This does one quarter of the job, but I'm not going to give you any more help than that; if you can work out why I said that, then you should be able to fill in the rest of the code.
if ((i ^ (i >> (5 - 2))) & (1 >> 2))
i ^= (1 << 2) | (1 << 5);
Essentially you need to reverse the bit ordering.
We're not going to solve this for you.. but here's a hint:
What if you had a 2-bit value. How would you reverse these bits?
A simple swap would work, right? Think about how to code this swap with operators that are available to you.
Now let's say you had a 4-bit value. How would you reverse these bits?
Could you split it into two 2-bit values, reverse each one, and then swap them? Would that give you the right result? Now code this.
Generalizing that solution to the 8-bit value should be trivial now.
Good luck!
I am trying to learn C programming, and I was studying some source codes and there are some things I didn't understand, especially regarding Bitwise Operators. I read some sites on this, and I kinda got an idea on what they do, but when I went back to look at this codes, I could not understand why and how where they used.
My first question is not related to bitwise operators but rather some ascii magic:
Can somebody explain to me how the following code works?
char a = 3;
int x = a - '0';
I understand this is done to convert a char into an int, however I don't understand the logic behind it. Why/How does it work?
Now, Regarding Bitwise operators, I feel really lost here.
What does this code do?
if (~pointer->intX & (1 << i)) { c++; n = i; }
I read somewhere that ~ inverts bits, but I fail to see what this statement is doing and why is it doing that.
Same with this line:
row.data = ~(1 << i);
Other question:
if (x != a)
{
ret |= ROW;
}
What exactly is the |= operator doing? From what I read, |= is OR but i don't quite understand what is this statement doing.
Is there any way of rewriting this code to make it easier to understands so that it doesn't use this bitwise operators? I find them very confusing to understand, so hopefully somebody will point me in the right direction on understanding how they work better!
I have a much better understanding of bitwise operators now and the whole code makes much more sense now.
One last thing: appartenly nobody responded if there would be a "cleaner" way for rewriting this code in a way that its easier to understand and maybe not at "bitlevel". Any ideas?
This will produce junk:
char a = 3;
int x = a - '0';
This is different - note the quotes:
char a = '3';
int x = a - '0';
The char datatype stores a number that identifiers a character. The characters for the digits 0 through 9 are all next to each other in the character code list, so if you subtract the code for '0' from the code for '9', you get the answer 9. So this will turn a digit character code into the integer value of the digit.
(~pointer->intX & (1 << i))
That will be interpreted by the if statement as true if it's non-zero. There are three different bitwise operators being used.
The ~ operator flips all the bits in the number, so if pointer->intX was 01101010, then ~pointer->intX will be 10010101. (Note that throughout, I'm illustrating the contents of a byte. If it was a 32-bit integer, I'd have to write 32 digits of 1s and 0s).
The & operator combines two numbers into one number, by dealing with each bit separately. The resulting bit is only 1 if both the input bits are 1. So if the left side is 00101001 and the right side is 00001011, the result will be 00001001.
Finally, << means left shift. If you start with 00000001 and left shift it by three places, you'll have 00001000. So the expression (1 << i) produces a value where bit i is switched on, and the others are all switch off.
Putting it all together, it tests if bit i is switched off (zero) in pointer->intX.
So you may be able to figure out what ~(1 << i) does. If i is 4, the thing in brackets will be 00010000, and so the whole thing will be 11101111.
ret |= ROW;
That one is equivalent to:
ret = ret | ROW;
The | operator is like & except that the resulting bit is 1 if either of the input bits is 1. So if ret is 00100000 and ROW is 00000010, the result will be 00100010.
ret |= ROW;
is equivalent to
ret = ret | ROW;
For char a = 3; int x = a - '0'; I think you meant char a = '3'; int x = a - '0';. It's easy enough to understand if you realize that in ASCII the numbers come in order, like '0', '1', '2', ... So if '0' is 48 and '1' is 49, then '1' - '0' is 1.
For bitwise operations, they are hard to grasp until you start looking at bits. When you view these operations on binary numbers then you can see exactly how they work...
010 & 111 = 010
010 | 111 = 111
010 ^ 111 = 101
~010 = 101
I think you probably have a typo, and meant:
char a = '3';
The reason this works is that all the numbers come in order, and '0' is the first. Obviously, '0' - '0' = 0. '1' - '0' = 1, since the character value for '1' is one greater than the character value for '0'. Etc.
1) A char is really just a 8-bit integer. '0' == 48, and all that that implies.
2) (~(pointer->intX) & (1 << i)) evalutates whether the 'i'th bit (from the right) in the intX member of whatever pointer points to is not set. The ~ inverts the bits, so all the 0s become 1s and vice versa, then the 1 << i puts a single 1 in the desired location, & combines the two values so that only the desired bit is kept, and the whole thing evalutes to true if that bit was 0 to begin with.
3) | is bitwise or. It takes each bit in both operands and performs a logical OR, producing a result where each bit is set if either operand had that bit set. 0b11000000 | 0b00000011 == 0b11000011. |= is an assignment operator, in the same way that a+=b means a=a+b, a|=b means a=a|b.
Not using bitwise operators CAN make things easier to read in some cases, but it will usually also make your code significantly slower without strong compiler optimization.
The subtraction trick you reference works because ASCII numbers are arranged in ascending order, starting with zero. So if ASCII '0' is a value of 48 (and it is), then '1' is a value of 49, '2' is 50, etc. Therefore ASCII('1') - ASCII('0') = 49 - 48 = 1.
As far as bitwise operators go, they allow you to perform bit-level operations on variables.
Let's break down your example:
(1 << i) -- this is left-shifting the constant 1 by i bits. So if i=0, the result is decimal 1. If i = 1, it shifts the bit one to the left, backfilling with zeros, yielding binary 0010, or decimal 2. If i = 2, you shift the bit two to the left, backfilling with zeros, yielding binary 0100 or decimal 4, etc.
~pointer->intX -- this is taking the value of the intX member of pointer and inverting its bits, setting all zeros to ones and vice versa.
& -- the ampersand operator does a bitwise AND comparison. The results of this will be 1 wherever both the left and right side of the expression are 1, and 0 otherwise.
So the test will succeed if pointer->intX has a 0 bit at the ith position from the right.
Also, |= means to do a bitwise OR comparison and assign the result to the left side of the expression. The result of a bitwise OR is 1 for every bit where the corresponding left or right side bit is 1,
Single quotes are used to indicate that a single char is used. '0' therefore is the char '0', which has the ASCII-Code 48.
3-'0'=3-48
'1<<i' shifts 1 i places to the left, therefore only the ith bit from the right is 1.
~pointer->intX negates the field intX, so the logical AND returns a true value (not 0) when intX has every bit except for the ith bit from the right isn't set.
char a = '3';
int x = a - '0';
you had a typo here (notice the 's around the 3), this assigns the ascii value of the character 3, to the char variable, then the next line takes '3' - '0' and assigns it to x, because of the way ascii values work, x will then be equal to 3 (integer value)
In the first comparison, I've never seen ~ being used on a pointer that way before, another typo maybe? If I were to read out the following code:
(~pointer->intX & (1 << i))
I would say "(the value intX dereferenced from pointer) AND (1 left shifted i times)"
1 << i is a quick way of multiplying 1 by a power of 2, ie if i is 3, then 1 << 3 == 8
In this case, I have no clue why you would invert the bits of the pointer..
In the 2nd comparison, x |= y is the same as x = x | y
I'm assuming you mean char a='3'; for the first line of code (otherwise you get a rather strange answer). The basic principal is that ASCII codes for digits are sequential, i.e. the code for '0'=48, the code for '1'=49, and so on. Subtracting '0' simply converts from the ASCII code to the actual digit, so e.g. '3' - '0' = 3, and so on. Note that this will only work if the character you're subtracting '0' from is an actual digit - otherwise the result will have little meaning.
a. Without context the "why" of this code is impossible to say. As for what it's doing, it appears that the if statement evaluates as true when bit i of pointer->intX is not set, i.e. that particular bit is a 0. I believe the & operator gets executed before the ~ operator, as the ~ operator has very low precedence. The code could make better use of parentheses to make the intended order of operations clearer. In this case, the order of operations might not matter though - I believe the result is the same either way.
b. This is simply creating a number with all bits EXCEPT bit i set to 1. A convenient way of creating a mask for bit i is to use the expression (1<<i).
The bitwise OR operation in this case is used to set the bits specified by the ROW constant to 1. If these bits are not set, it sets them; if they're already set it has no effect.
1) Can somebody explain to me how the following code works? char a = 3; int x = a - '0';
I undertand this is done to convert a char into an int, however I don't understand the logic behind it. Why/How does it work?
Sure. variable a is of type char, and by putting single quotes around 0 that is causing C to view it as a char as well. Finally, the whole statement is automagically typecast to its integer equivalent, because x is defined as an integer.
2) Now, Regarding Bitwise operators, I feel really lost here.
--- What does this code do? if (~pointer->intX & (1 << i)) { c++; n = i; } I read somewhere that ~ inverts bits, but I fail to see what this statement is doing and why is it doing that.
(~pointer->intX & (1 << i)) is saying:
negate intX, and AND it with a 1 shifted left by i bits
so, what you're getting, if intX = 1011, and i = 2, equates to
(0100 & 0100)
-negate 1011 = 0100
-(1 << 2) = 0100
0100 & 0100 = 1 :)
then, if the AND operation returns a 1 (which, in my example, it does)
{ c++; n = i; }
so, increment c by 1, and set variable n to be i
Same with this line: row.data = ~(1 << i);
Same principle here.
Shift a 1 to the left by i places, and negate.
So, if i = 2 again
(1 << 2) = 0100
~(0100) = 1011
**--- Other question:
if (x != a) { ret |= ROW; }
What exacly is the |= operator doing? From what I read, |= is OR but i don't quite understand what is this statement doing.**
if (x != a) (hopefully this is apparent to you....if variable x does not equal variable a)
ret |= ROW;
equates to
ret = ret | ROW;
which means, binary OR ret with ROW
For examples of exactly what AND and OR operations accomplish, you should have a decent understanding of binary logic.
Check wikipedia for truth tables...ie
Bitwise operations