Can someone explain to me the reason why someone would want use bitwise comparison?
example:
int f(int x) {
return x & (x-1);
}
int main(){
printf("F(10) = %d", f(10));
}
This is what I really want to know: "Why check for common set bits"
x is any positive number.
Bitwise operations are used for three reasons:
You can use the least possible space to store information
You can compare/modify an entire register (e.g. 32, 64, or 128 bits depending on your processor) in a single CPU instruction, usually taking a single clock cycle. That means you can do a lot of work (of certain types) blindingly fast compared to regular arithmetic.
It's cool, fun and interesting. Programmers like these things, and they can often be the differentiator when there is no difference between techniques in terms of efficiency/performance.
You can use this for all kinds of very handy things. For example, in my database I can store a lot of true/false information about my customers in a tiny space (a single byte can store 8 different true/false facts) and then use '&' operations to query their status:
Is my customer Male and Single and a Smoker?
if (customerFlags & (maleFlag | singleFlag | smokerFlag) ==
(maleFlag | singleFlag | smokerFlag))
Is my customer (any combination of) Male Or Single Or a Smoker?
if (customerFlags & (maleFlag | singleFlag | smokerFlag) != 0)
Is my customer not Male and not Single and not a Smoker)?
if (customerFlags & (maleFlag | singleFlag | smokerFlag) == 0)
Aside from just "checking for common bits", you can also do:
Certain arithmetic, e.g. value & 15 is a much faster equivalent of value % 16. This only works for certain numbers, but if you can use it, it can be a great optimisation.
Data packing/unpacking. e.g. a colour is often expressed as a 32-bit integer that contains Alpha, Red, Green and Blue byte values. The Red value might be extracted with an expression like red = (value >> 16) & 255; (shift the value down 16 bit positions and then carve off the bottom byte)
Data manipulation and swizzling. Some clever tricks can be achieved with bitwise operations. For example, swapping two integer values without needing to use a third temporary variable, or converting ARGB colour values into another format (e.g RGBA or BGRA)
The Ur-example is "testing if a number is even or odd":
unsigned int number = ...;
bool isOdd = (0 != (number & 1));
More complex uses include bitmasks (multiple boolean values in a single integer, each one taking up one bit of space) and encryption/hashing (which frequently involve bit shifting, XOR, etc.)
The example you've given is kinda odd, but I'll use bitwise comparisons all the time in embedded code.
I'll often have code that looks like the following:
volatile uint32_t *flags = 0x000A000;
bool flagA = *flags & 0x1;
bool flagB = *flags & 0x2;
bool flagC = *flags & 0x4;
It's not a bitwise comparison. It doesn't return a boolean.
Bitwise operators are used to read and modify individual bits of a number.
n & 0x8 // Peek at bit3
n |= 0x8 // Set bit3
n &= ~0x8 // Clear bit3
n ^= 0x8 // Toggle bit3
Bits are used in order to save space. 8 chars takes a lot more memory than 8 bits in a char.
The following example gets the range of an IP subnet using given an IP address of the subnet and the subnet mask of the subnet.
uint32_t mask = (((255 << 8) | 255) << 8) | 255) << 8) | 255;
uint32_t ip = (((192 << 8) | 168) << 8) | 3) << 8) | 4;
uint32_t first = ip & mask;
uint32_t last = ip | ~mask;
e.g. if you have a number of status flags in order to save space you may want to put each flag as a bit.
so x, if declared as a byte, would have 8 flags.
I think you mean bitwise combination (in your case a bitwise AND operation). This is a very common operation in those cases where the byte, word or dword value is handled as a collection of bits, eg status information, eg in SCADA or control programs.
Your example tests whether x has at most 1 bit set. f returns 0 if x is a power of 2 and non-zero if it is not.
Your particular example tests if two consecutive bits in the binary representation are 1.
Related
Take macro:
GPIOxMODE(gpio,mode,port) ( GPIO##gpio->MODER = ((GPIO##gpio->MODER & ~((uint32_t)GPIO2BITMASK << (port*2))) | (mode << (port * 2))) )
Assuming that the reset value of the register is 0xFFFF.FFFF, I want to set a 2 bit width to an arbitrary value. This was written for an STM32
MCU that has 15 pins per port. GPIO2BITMASK is defined as 0x3. Is there a better way for clearing and setting a random 2 bits in anywhere in the
32-bit wide register.
Valid range for port 0 - 15
Valid range for mode 0 - 3
The method I came up with is to bit shift the mask, invert it, logically AND it with the existing register value, logically OR the result with a bit shifted new value.
I am looking to combine the mask and new value to reduce the number of logical operations bit shift operations. The goal is also keep the process generic enough so that I can use for bit operations of 1,2,3 or 4 bit widths.
Is there a better way?
In the long and sort of it, is there a better way is really an opened question. I am looking specifically for a method that will reduce the number of logical operations and bit shift operations, while being a simple one lined statement.
The answer is NO.
You MUST do reset/set to ensure that the bit field you are writing to has the desired value.
The answers received can be better (in a matter of opinion/preference/philosophy/practice) in that they aren't necessary a macros and have have parameter checking. Also pit falls of this style have been pointed out in both the comments and responses.
This kind of macros should be avoided as a plaque for many reasons:
They are not debuggable
They are hard to find error prone
and many other reasons
The same result you can archive using inline functions. The resulting code will be the same effective
static inline __attribute__((always_inline)) void GPIOMODE(GPIO_TypeDef *gpio, unsigned mode, unsigned pin)
{
gpio -> MODER &= ~(GPIO_MODER_MODE0_Msk << (pin * 2));
gpio -> MODER |= mode << (pin * 2);
}
but if you love macros
#define GPIOxMODE(gpio,mode,port) {volatile uint32_t *mdr = &GPIO##gpio->MODER; *mdr &= ~(GPIO_MODER_MODE0_Msk << (port*2)); *mdr |= mode << (port * 2);}
I am looking to combine the mask and new value to reduce the number of
logical operations bit shift operations.
you cant. You need to reset and then set the bits.
The method I came up with is to bit shift the mask, invert it,
logically AND it with the existing register value, logically OR the
result with a bit shifted new value.
That or an equivalent is the way to do it.
I am looking to combine the mask and new value to reduce the number of
logical operations bit shift operations. The goal is also keep the
process generic enough so that I can use for bit operations of 1,2,3
or 4 bit widths.
Is there a better way?
You must accomplish two basic objectives:
ensure that the bits that should be off in the affected range are in fact off, and
ensure that the bits that should be on in the affected range are in fact on.
In the general case, those require two separate operations: a bitwise AND to force bits off, and a bitwise OR (or XOR, if the bits are first cleared) to turn the wanted bits on. There may be ways to shortcut for specific cases of original and target values, but if you want something general-purpose, as you say, then your options are limited.
Personally, though, I think I would be inclined to build it from multiple pieces, separating the GPIO selection from the actual computation. At minimum, you can separate out a generic macro for setting a range of bits:
#define SETBITS32(x,bits,offset,mask) ((((uint32_t)(x)) & ~(((uint32_t)(mask)) << (offset))) | (((uint32_t)(bits)) << (offset)))
#define GPIOxMODE(gpio,mode,port) (GPIO##gpio->MODER = SETBITS32(GPIO##gpio->MODER, mode, port * 2, GPIO2BITMASK)
But do note that there appears to be no good way to avoid such a macro evaluating some of its arguments more than once. It might therefore be safer to write SETBITS32 as a function instead. The compiler will probably inline such a function in any case, but you can maximize the likelihood of that by declaring it static and inline:
static inline uint32_t SETBITS32(uint32_t x, uint32_t bits, unsigned offset, uint32_t mask) {
return x & ~(mask << offset) | (bits << offset);
}
That's easier to read, too, though it, like the macro, does assume that bits has no set bits outside the mask region.
Of course there are other, similar formulations. For instance, if you do not need to support discontinuous bit ranges, you might specify a bit count instead of a bit mask. This alternative does that, protects against the user providing bits outside the specified range, and also has some parameter validation:
static inline uint32_t set_bitrange_32(uint32_t x, uint32_t bits, unsigned width,
unsigned offset) {
if (width + offset > 32) {
// error: invalid parameters
return x;
} else if (width == 0) {
return x;
}
uint32_t mask = ~(uint32_t)0 >> (32 - width);
return x & ~(mask << offset) | ((bits & mask) << offset);
}
I'm querying an ADXL362 Digital Output MEMS Accelerometer for its axis data which it holds as two 8 bit registers which combine to give a 12 bit value and I'm trying to figure out how to combine those values. I've never been good at bitwise manipulation so any help would be greatly appreciated. I would imagine it is something like this:
number = Z_data_H << 8 | Z_data_L;
number = (number & ~(1<<13)) | (0<<13);
number = (number & ~(1<<14)) | (0<<14);
number = (number & ~(1<<15)) | (0<<15);
number = (number & ~(1<<16)) | (0<<16);
ADXL362 data sheet (page 26)
Z axis data register
Your first line should be what you need:
int16_t number;
number = (Z_data_H << 8) | Z_data_L;
The sign-extension bits mean that you can read the value as if it was a 16-bit signed integer. The value will simply never be outside the range of a 12-bit integer. It's important that you leave those bits intact in order to handle negative values correctly.
You just have to do:
signed short number;
number = Z_data_H << 8 | Z_data_L;
The shift left by 8 bit combined with the lower bits you already
had figured out are combining the 2 bytes correctly. Just use the appropriate data size to have the C code recoginize the sign of the 12 bit number correctly.
Note that short not necessarily refers to a 16bit value, depending on your compiler and architecture - so, you might want to attempt to that.
This question already has answers here:
Real world use cases of bitwise operators [closed]
(41 answers)
Closed 6 years ago.
I am new to bitwise operators.
I understand how the logic functions work to get the final result. For example, when you bitwise AND two numbers, the final result is going to be the AND of those two numbers (1 & 0 = 0; 1 & 1 = 1; 0 & 0 = 0). Same with OR, XOR, and NOT.
What I don't understand is their application. I tried looking everywhere and most of them just explain how bitwise operations work. Of all the bitwise operators I only understand the application of shift operators (multiplication and division). I also came across masking. I understand that masking is done using bitwise AND but what exactly is its purpose and where and how can I use it?
Can you elaborate on how I can use masking? Are there similar uses for OR and XOR?
The low-level use case for the bitwise operators is to perform base 2 math. There is the well known trick to test if a number is a power of 2:
if ((x & (x - 1)) == 0) {
printf("%d is a power of 2\n", x);
}
But, it can also serve a higher level function: set manipulation. You can think of a collection of bits as a set. To explain, let each bit in a byte to represent 8 distinct items, say the planets in our solar system (Pluto is no longer considered a planet, so 8 bits are enough!):
#define Mercury (1 << 0)
#define Venus (1 << 1)
#define Earth (1 << 2)
#define Mars (1 << 3)
#define Jupiter (1 << 4)
#define Saturn (1 << 5)
#define Uranus (1 << 6)
#define Neptune (1 << 7)
Then, we can form a collection of planets (a subset) like using |:
unsigned char Giants = (Jupiter|Saturn|Uranus|Neptune);
unsigned char Visited = (Venus|Earth|Mars);
unsigned char BeyondTheBelt = (Jupiter|Saturn|Uranus|Neptune);
unsigned char All = (Mercury|Venus|Earth|Mars|Jupiter|Saturn|Uranus|Neptune);
Now, you can use a & to test if two sets have an intersection:
if (Visited & Giants) {
puts("we might be giants");
}
The ^ operation is often used to see what is different between two sets (the union of the sets minus their intersection):
if (Giants ^ BeyondTheBelt) {
puts("there are non-giants out there");
}
So, think of | as union, & as intersection, and ^ as union minus the intersection.
Once you buy into the idea of bits representing a set, then the bitwise operations are naturally there to help manipulate those sets.
One application of bitwise ANDs is checking if a single bit is set in a byte. This is useful in networked communication, where protocol headers attempt to pack as much information into the smallest area as is possible in an effort to reduce overhead.
For example, the IPv4 header utilizes the first 3 bits of the 6th byte to tell whether the given IP packet can be fragmented, and if so whether to expect more fragments of the given packet to follow. If these fields were the size of ints (1 byte) instead, each IP packet would be 21 bits larger than necessary. This translates to a huge amount of unnecessary data through the internet every day.
To retrieve these 3 bits, a bitwise AND could be used along side a bit mask to determine if they are set.
char mymask = 0x80;
if(mymask & (ipheader + 48) == mymask)
//the second bit of the 6th byte of the ip header is set
Small sets, as has been mentioned. You can do a surprisingly large number of operations quickly, intersection and union and (symmetric) difference are obviously trivial, but for example you can also efficiently:
get the lowest item in the set with x & -x
remove the lowest item from the set with x & (x - 1)
add all items smaller than the smallest present item
add all items higher than the smallest present item
calculate their cardinality (though the algorithm is nontrivial)
permute the set in some ways, that is, change the indexes of the items (not all permutations are equally efficient)
calculate the lexicographically next set that contains as many items (Gosper's Hack)
1 and 2 and their variations can be used to build efficient graph algorithms on small graphs, for example see algorithm R in The Art of Computer Programming 4A.
Other applications of bitwise operations include, but are not limited to,
Bitboards, important in many board games. Chess without bitboards is like Christmas without Santa. Not only is it a space-efficient representation, you can do non-trivial computations directly with the bitboard (see Hyperbola Quintessence)
sideways heaps, and their application in finding the Nearest Common Ancestor and computing Range Minimum Queries.
efficient cycle-detection (Gosper's Loop Detection, found in HAKMEM)
adding offsets to Z-curve addresses without deconstructing and reconstructing them (see Tesseral Arithmetic)
These uses are more powerful, but also advanced, rare, and very specific. They show, however, that bitwise operations are not just a cute toy left over from the old low-level days.
Example 1
If you have 10 booleans that "work together" you can do simplify your code a lot.
int B1 = 0x01;
int B2 = 0x02;
int B10 = 0x0A;
int someValue = get_a_value_from_somewhere();
if (someValue & (B1 + B10)) {
// B1 and B10 are set
}
Example 2
Interfacing with hardware. An address on the hardware may need bit level access to control the interface. e.g. an overflow bit on a buffer or a status byte that can tell you the status of 8 different things. Using bit masking you can get down the the actual bit of info you need.
if (register & 0x80) {
// top bit in the byte is set which may have special meaning.
}
This is really just a specialized case of example 1.
Bitwise operators are particularly useful in systems with limited resources as each bit can encode a boolean. Using many chars for flags is wasteful as each takes one byte of space (when they could be storing 8 flags each).
Commonly microcontrollers have C interfaces for their IO ports in which each bit controls 1 of 8 ports. Without bitwise operators these would be quite difficult to control.
Regarding masking, it is common to use both & and |:
x & 0x0F //ensures the 4 high bits are 0
x | 0x0F //ensures the 4 low bits are 1
In microcontroller applications, you can utilize bitwise to switch between ports. In the below picture, if we would like to turn on a single port while turning off the rest, then the following code can be used.
void main()
{
unsigned char ON = 1;
TRISB=0;
PORTB=0;
while(1){
PORTB = ON;
delay_ms(200);
ON = ON << 1;
if(ON == 0) ON=1;
}
}
I'm reading two registers from microcontroller. One have 4-bit MSB (First 4-bits has some other things) and another 8-bit LSB. I want to convert it into one 12-bit uint (16 bit to be precise). So far I made it like that:
UINT16 x;
UINT8 RegValue = 0;
UINT8 RegValue1 = 0;
ReadRegister(Register01, &RegValue1);
ReadRegister(Register02, &RegValue2);
x = RegValue1 & 0x000F;
x = x << 8;
x = x | RegValue2 & 0x00FF;
is there any better way to do that?
/* To be more precise ReadRegister is I2C communication to another ADC. Register01 and Register02 are different addresses. RegValue1 is 8 bit but only 4 LSB are needed and concatenate to RegValue (4-LSB of RegValue1 and all 8-bits of RegValue). */
If you know the endianness of your machine, you can read the bytes
directly into x like this:
ReadRegister(Register01, (UINT8*)&x + 1);
ReadRegister(Register02, (UINT8*)&x);
x &= 0xfff;
Note that this is not portable and the performance gain (if any) will
likely be small.
The RegValue & 0x00FF mask is unnecessary since RegValue is already 8 bit.
Breaking it down into three statements may be good for clarity, but this expression is probably simple enough to implement in one statement:
x = ((RegValue1 & 0x0Fu) << 8u) | RegValue ;
The use of an unsigned literal (0x0Fu) makes little difference but emphasises that we are dealing with unsigned 8-bit data. It is in fact an unsigned int even with only two digits, but again this emphasises to the reader perhaps that we are only dealing with 8 bits, and is purely stylistic rather than semantic. In C there is no 8-bit literal constant type (though in C++ '\x0f' has type char). You can force better type agreement as follows:
#define LS4BITMASK ((UINT8)0x0fu)
x = ((RegValue1 & LS4BITMASK) << 8u) | RegValue ;
The macro merely avoids repetition and clutter in the expression.
None of the above is necessarily "better" than your original code in terms of performance or actual generated code, and is largely a matter of preference or local coding standards or practices.
If the registers are adjacent to each other, they will most likley also be in the correct order with respect to target endianness. That being the case they can be read as a single 16 bit register and masked accordingly, assuming that Register01 is the lower address value:
ReadRegister16(Register01, &x ) ;
x &= 0x0fffu ;
Of course I have invented here the ReadRegister16() function, but if the registers are memory mapped, and Register01 is simply an address then this may simply be:
UINT16 x = *Register01 ;
x &= 0x0fffu ;
I'm trying to implement a data compression idea I've had, and since I'm imagining running it against a large corpus of test data, I had thought to code it in C (I mostly have experience in scripting languages like Ruby and Tcl.)
Looking through the O'Reilly 'cow' books on C, I realize that I can't simply index the bits of a simple 'char' or 'int' type variable as I'd like to to do bitwise comparisons and operators.
Am I correct in this perception? Is it reasonable for me to use an enumerated type for representing a bit (and make an array of these, and writing functions to convert to and from char)? If so, is such a type and functions defined in a standard library already somewhere? Are there other (better?) approaches? Is there some example code somewhere that someone could point me to?
Thanks -
Following on from what Kyle has said, you can use a macro to do the hard work for you.
It is possible.
To set the nth bit, use OR:
x |= (1 << 5); // sets the 6th-from
right
To clear a bit, use AND:
x &= ~(1 << 5); // clears
6th-from-right
To flip a bit, use XOR:
x ^= (1 << 5); // flips 6th-from-right
Or...
#define GetBit(var, bit) ((var & (1 << bit)) != 0) // Returns true / false if bit is set
#define SetBit(var, bit) (var |= (1 << bit))
#define FlipBit(var, bit) (var ^= (1 << bit))
Then you can use it in code like:
int myVar = 0;
SetBit(myVar, 5);
if (GetBit(myVar, 5))
{
// Do something
}
It is possible.
To set the nth bit, use OR:
x |= (1 << 5); // sets the 5th-from right
To clear a bit, use AND:
x &= ~(1 << 5); // clears 5th-from-right
To flip a bit, use XOR:
x ^= (1 << 5); // flips 5th-from-right
To get the value of a bit use shift and AND:
(x & (1 << 5)) >> 5 // gets the value (0 or 1) of the 5th-from-right
note: the shift right 5 is to ensure the value is either 0 or 1. If you're just interested in 0/not 0, you can get by without the shift.
Have a look at the answers to this question.
Theory
There is no C syntax for accessing or setting the n-th bit of a built-in datatype (e.g. a 'char'). However, you can access bits using a logical AND operation, and set bits using a logical OR operation.
As an example, say that you have a variable that holds 1101 and you want to check the 2nd bit from the left. Simply perform a logical AND with 0100:
1101
0100
---- AND
0100
If the result is non-zero, then the 2nd bit must have been set; otherwise is was not set.
If you want to set the 3rd bit from the left, then perform a logical OR with 0010:
1101
0010
---- OR
1111
You can use the C operators && (for AND) and || (for OR) to perform these tasks. You will need to construct the bit access patterns (the 0100 and 0010 in the above examples) yourself. The trick is to remember that the least significant bit (LSB) counts 1s, the next LSB counts 2s, then 4s etc. So, the bit access pattern for the n-th LSB (starting at 0) is simply the value of 2^n. The easiest way to compute this in C is to shift the binary value 0001 (in this four bit example) to the left by the required number of places. As this value is always equal to 1 in unsigned integer-like quantities, this is just '1 << n'
Example
unsigned char myVal = 0x65; /* in hex; this is 01100101 in binary. */
/* Q: is the 3-rd least significant bit set (again, the LSB is the 0th bit)? */
unsigned char pattern = 1;
pattern <<= 3; /* Shift pattern left by three places.*/
if(myVal && (char)(1<<3)) {printf("Yes!\n");} /* Perform the test. */
/* Set the most significant bit. */
myVal |= (char)(1<<7);
This example hasn't been tested, but should serve to illustrate the general idea.
To query state of bit with specific index:
int index_state = variable & ( 1 << bit_index );
To set bit:
varabile |= 1 << bit_index;
To restart bit:
variable &= ~( 1 << bit_index );
Try using bitfields. Be careful the implementation can vary by compiler.
http://publications.gbdirect.co.uk/c_book/chapter6/bitfields.html
IF you want to index a bit you could:
bit = (char & 0xF0) >> 7;
gets the msb of a char. You could even leave out the right shift and do a test on 0.
bit = char & 0xF0;
if the bit is set the result will be > 0;
obviousuly, you need to change the mask to get different bits (NB: the 0xF is the bit mask if it is unclear). It is possible to define numerous masks e.g.
#define BIT_0 0x1 // or 1 << 0
#define BIT_1 0x2 // or 1 << 1
#define BIT_2 0x4 // or 1 << 2
#define BIT_3 0x8 // or 1 << 3
etc...
This gives you:
bit = char & BIT_1;
You can use these definitions in the above code to sucessfully index a bit within either a macro or a function.
To set a bit:
char |= BIT_2;
To clear a bit:
char &= ~BIT_3
To toggle a bit
char ^= BIT_4
This help?
Individual bits can be indexed as follows.
Define a struct like this one:
struct
{
unsigned bit0 : 1;
unsigned bit1 : 1;
unsigned bit2 : 1;
unsigned bit3 : 1;
unsigned reserved : 28;
} bitPattern;
Now if I want to know the individual bit values of a var named "value", do the following:
CopyMemory( &input, &value, sizeof(value) );
To see if bit 2 is high or low:
int state = bitPattern.bit2;
Hope this helps.
There is a standard library container for bits: std::vector. It is specialised in the library to be space efficient. There is also a boost dynamic_bitset class.
These will let you perform operations on a set of boolean values, using one bit per value of underlying storage.
Boost dynamic bitset documentation
For the STL documentation, see your compiler documentation.
Of course, you can also address the individual bits in other integral types by hand. If you do that, you should use unsigned types so that you don't get undefined behaviour if decide to do a right shift on a value with the high bit set. However, it sounds like you want the containers.
To the commenter who claimed this takes 32x more space than necessary: boost::dynamic_bitset and vector are specialised to use one bit per entry, and so there is not a space penalty, assuming that you actually want more than the number of bits in a primitive type. These classes allow you to address individual bits in a large container with efficient underlying storage. If you just want (say) 32 bits, by all means, use an int. If you want some large number of bits, you can use a library container.