I want to know, how could I extend the most significant bit (bit 23) into bits 24 through 31? How could I do that in C code? I am using C code to program Nios II.
I was thinking of using bit shifting operation but not knowing in details how by using bit shifting operation, the above could be achieved, any link or resource is much appreciated.
Thank you in advance.
As Carl said, right shift if implementation defined. You can use other binary operators that will always work:
if (0 != (0x00800000 & x)) //test if bit 23 is set
{
x |= 0xFF000000; //set bits 24-31
}
else
{
x &= 0x00FFFFFF; //clear bits 24-31
}
The C right-shift operator has implementation-defined behaviour when right-shifting. Since Nios II has an arithmetic right-shift instruction, you can likely simply do:
x = (x << 8) >> 8;
Double check the output assembly to be sure it uses an instruction from the sra family.
A variation on #IronMensan which relies on the reasonable assumption that the integer being modified is 32 bits.
The following only affects bit 24-31, even if the integer is wider.
#define Mask2431 (0xFF000000)
#define Bit23 (0x800000)
some_int |= Mask2431;
if (!(some_int & Bit23))
some_int ^= Mask2431;
The following affects bit 24 and all higher even when using wider than a 32-bit integer:
#define Mask24 (0xFFFFFF)
#define Bit23 (0x800000)
some_int &= Mask24;
if (some_int & Bit23)
some_int = ~some_int ^ Mask24;
Related
I'm trying to get the most significant bit of an unsigned 8-bit type in C.
This is what I'm trying to do right now:
uint8_t *var = ...;
...
(*var >> 6) & 1
Is this right? If it's not, what would be?
To get the most significant bit from a memory pointed to by uint8_t pointer, you need to shift by 7 bits.
(*var >> 7) & 1
The most standard/correct way of masking bits is to use a readable bit mask of the form 1u << bit. Any C programmer spotting 1u << n in code will know that it is a bit mask - so it is self-documenting code.
So if you want bit number 7, you would write
*var & (1u << 7)
The u suffix is important for rugged code, since you want to avoid accidental implicit promotions to signed types.
Another option is to simply apply a bit mask and check the resulting value:
*var & 0x80u // 1000 0000
I have a sensor which gives its output in three bytes. I read it like this:
unsigned char byte0,byte1,byte2;
byte0=readRegister(0x25);
byte1=readRegister(0x26);
byte2=readRegister(0x27);
Now I want these three bytes merged into one number:
int value;
value=byte0 + (byte1 << 8) + (byte2 << 16);
it gives me values from 0 to 16,777,215 but I'm expecting values from -8,388,608 to 8,388,607. I though that int was already signed by its implementation. Even if I try define it like signed int value; it still gives me only positive numbers. So I guess my question is how to convert int to its two's complement?
Thanks!
What you need to perform is called sign extension. You have 24 significant bits but want 32 significant bits (note that you assume int to be 32-bit wide, which is not always true; you'd better use type int32_t defined in stdint.h). Missing 8 top bits should be either all zeroes for positive values or all ones for negative. It is defined by the most significant bit of the 24 bit value.
int32_t value;
uint8_t extension = byte2 & 0x80 ? 0xff:00; /* checks bit 7 */
value = (int32_t)byte0 | ((int32_t)byte1 << 8) | ((int32_t)byte2 << 16) | ((int32_t)extension << 24);
EDIT: Note that you cannot shift an 8 bit value by 8 or more bits, it is undefined behavior. You'll have to cast it to a wider type first.
#include <stdint.h>
uint8_t byte0,byte1,byte2;
int32_t answer;
// assuming reg 0x25 is the signed MSB of the number
// but you need to read unsigned for some reason
byte0=readRegister(0x25);
byte1=readRegister(0x26);
byte2=readRegister(0x27);
// so the trick is you need to get the byte to sign extend to 32 bits
// so force it signed then cast it up
answer = (int32_t)((int8_t)byte0); // this should sign extend the number
answer <<= 8;
answer |= (int32_t)byte1; // this should just make 8 bit field, not extended
answer <<= 8;
answer |= (int32_t)byte2;
This should also work
answer = (((int32_t)((int8_t)byte0))<<16) + (((int32_t)byte1)<< 8) + byte2;
I may be overly aggressive with parentheses but I never trust myself with shift operators :)
I encountered this piece of C code that's existing. I am struggling to understand it.
I supposidly reads a 4 byte unsigned value passed in a buffer (in little endian format) into a variable of type "long".
This code runs on a 64 bit word size, little endian x86 machine - where sizeof(long) is 8 bytes.
My guess is that this code is intended to also run on a 32 bit x86 machine - so a variable of type long is used instead of int for sake of storing value from a four byte input data.
I am having some doubts and have put comments in the code to express what I understand, or what I don't :-)
Please answer questions below in that context
void read_Value_From_Four_Byte_Buff( char*input)
{
/* use long so on 32 bit machine, can still accommodate 4 bytes */
long intValueOfInput;
/* Bitwise and of input buffer's byte 0 with 0xFF gives MSB or LSB ?*/
/* This code seems to assume that assignment will store in rightmost byte - is that true on a x86 machine ?*/
intValueOfInput = 0xFF & input[0];
/*left shift byte-1 eight times, bitwise "or" places in 2nd byte frm right*/
intValueOfInput |= ((0xFF & input[1]) << 8);
/* similar left shift in mult. of 8 and bitwise "or" for next two bytes */
intValueOfInput |= ((0xFF & input[2]) << 16);
intValueOfInput |= ((0xFF & input[3]) << 24);
}
My questions
1) The input buffer is expected to be in "Little endian". But from code looks like assumption here is that it read in as Byte 0 = MSB, Byte 1, Byte 2, Byte 3= LSB. I thought so because code reads bytes starting from Byte 0, and subsequent bytes ( 1 onwards) are placed in the target variable after left shifting. Is that how it is or am I getting it wrong ?
2) I feel this is a convoluted way of doing things - is there a simpler alternative to copy value from 4 byte buffer into a long variable ?
3) Will the assumption "that this code will run on a 64 bit machine" will have any bearing on how easily I can do this alternatively? I mean is all this trouble to keep it agnostic to word size ( I assume its agnostic to word size now - not sure though) ?
Thanks for your enlightenment :-)
You have it backwards. When you left shift, you're putting into more significant bits. So (0xFF & input[3]) << 24) puts Byte 3 into the MSB.
This is the way to do it in standard C. POSIX has the function ntohl() that converts from network byte order to a native 32-bit integer, so this is usually used in Unix/Linux applications.
This will not work exactly the same on a 64-bit machine, unless you use unsigned long instead of long. As currently written, the highest bit of input[3] will be put into the sign bit of the result (assuming a twos-complement machine), so you can get negative results. If long is 64 bits, all the results will be positive.
The code you are using does indeed treat the input buffer as little endian. Look how it takes the first byte of the buffer and just assigns it to the variable without any shifting. If the first byte increases by 1, the value of your result increases by 1, so it is the least-significant byte (LSB). Left-shifting makes a byte more significant, not less. Left-shifting by 8 is generally the same as multiplying by 256.
I don't think you can get much simpler than this unless you use an external function, or make assumptions about the machine this code is running on, or invoke undefined behavior. In most instances, it would work to just write uint32_t x = *(uint32_t *)input; but this assumes your machine is little endian and I think it might be undefined behavior according to the C standard.
No, running on a 64-bit machine is not a problem. I recommend using types like uint32_t and int32_t to make it easier to reason about whether your code will work on different architectures. You just need to include the stdint.h header from C99 to use those types.
The right-hand side of the last line of this function might exhibit undefined behavior depending on the data in the input:
((0xFF & input[3]) << 24)
The problem is that (0xFF & input[3]) will be a signed int (because of integer promotion). The int will probably be 32-bit, and you are shifting it so far to the left that the resulting value might not be representable in an int. The C standard says this is undefined behavior, and you should really try to avoid that because it gives the compiler a license to do whatever it wants and you won't be able to predict the result.
A solution is to convert it from an int to a uint32_t before shifting it, using a cast.
Finally, the variable intValueOfInput is written to but never used. Shouldn't you return it or store it somewhere?
Taking all this into account, I would rewrite the function like this:
uint32_t read_value_from_four_byte_buff(char * input)
{
uint32_t x;
x = 0xFF & input[0];
x |= (0xFF & input[1]) << 8;
x |= (0xFF & input[2]) << 16;
x |= (uint32_t)(0xFF & input[3]) << 24;
return x;
}
From the code, Byte 0 is LSB, Byte 3 is MSB. But there are some typos. The lines should be
intValueOfInput |= ((0xFF & input[2]) << 16);
intValueOfInput |= ((0xFF & input[3]) << 24);
You can make the code shorter by dropping 0xFF but using the type "unsigned char" in the argument type.
To make the code shorter, you can do:
long intValueOfInput = 0;
for (int i = 0, shift = 0; i < 4; i++, shift += 8)
intValueOfInput |= ((unsigned char)input[i]) << shift;
Can someone please explain this function to me?
A mask with the least significant n bits set to 1.
Ex:
n = 6 --> 0x2F, n = 17 --> 0x1FFFF // I don't get these at all, especially how n = 6 --> 0x2F
Also, what is a mask?
The usual way is to take a 1, and shift it left n bits. That will give you something like: 00100000. Then subtract one from that, which will clear the bit that's set, and set all the less significant bits, so in this case we'd get: 00011111.
A mask is normally used with bitwise operations, especially and. You'd use the mask above to get the 5 least significant bits by themselves, isolated from anything else that might be present. This is especially common when dealing with hardware that will often have a single hardware register containing bits representing a number of entirely separate, unrelated quantities and/or flags.
A mask is a common term for an integer value that is bit-wise ANDed, ORed, XORed, etc with another integer value.
For example, if you want to extract the 8 least significant digits of an int variable, you do variable & 0xFF. 0xFF is a mask.
Likewise if you want to set bits 0 and 8, you do variable | 0x101, where 0x101 is a mask.
Or if you want to invert the same bits, you do variable ^ 0x101, where 0x101 is a mask.
To generate a mask for your case you should exploit the simple mathematical fact that if you add 1 to your mask (the mask having all its least significant bits set to 1 and the rest to 0), you get a value that is a power of 2.
So, if you generate the closest power of 2, then you can subtract 1 from it to get the mask.
Positive powers of 2 are easily generated with the left shift << operator in C.
Hence, 1 << n yields 2n. In binary it's 10...0 with n 0s.
(1 << n) - 1 will produce a mask with n lowest bits set to 1.
Now, you need to watch out for overflows in left shifts. In C (and in C++) you can't legally shift a variable left by as many bit positions as the variable has, so if ints are 32-bit, 1<<32 results in undefined behavior. Signed integer overflows should also be avoided, so you should use unsigned values, e.g. 1u << 31.
For both correctness and performance, the best way to accomplish this has changed since this question was asked back in 2012 due to the advent of BMI instructions in modern x86 processors, specifically BLSMSK.
Here's a good way of approaching this problem, while retaining backwards compatibility with older processors.
This method is correct, whereas the current top answers produce undefined behavior in edge cases.
Clang and GCC, when allowed to optimize using BMI instructions, will condense gen_mask() to just two ops. With supporting hardware, be sure to add compiler flags for BMI instructions:
-mbmi -mbmi2
#include <inttypes.h>
#include <stdio.h>
uint64_t gen_mask(const uint_fast8_t msb) {
const uint64_t src = (uint64_t)1 << msb;
return (src - 1) ^ src;
}
int main() {
uint_fast8_t msb;
for (msb = 0; msb < 64; ++msb) {
printf("%016" PRIx64 "\n", gen_mask(msb));
}
return 0;
}
First, for those who only want the code to create the mask:
uint64_t bits = 6;
uint64_t mask = ((uint64_t)1 << bits) - 1;
# Results in 0b111111 (or 0x03F)
Thanks to #Benni who asked about using bits = 64. If you need the code to support this value as well, you can use:
uint64_t bits = 6;
uint64_t mask = (bits < 64)
? ((uint64_t)1 << bits) - 1
: (uint64_t)0 - 1
For those who want to know what a mask is:
A mask is usually a name for value that we use to manipulate other values using bitwise operations such as AND, OR, XOR, etc.
Short masks are usually represented in binary, where we can explicitly see all the bits that are set to 1.
Longer masks are usually represented in hexadecimal, that is really easy to read once you get a hold of it.
You can read more about bitwise operations in C here.
I believe your first example should be 0x3f.
0x3f is hexadecimal notation for the number 63 which is 111111 in binary, so that last 6 bits (the least significant 6 bits) are set to 1.
The following little C program will calculate the correct mask:
#include <stdarg.h>
#include <stdio.h>
int mask_for_n_bits(int n)
{
int mask = 0;
for (int i = 0; i < n; ++i)
mask |= 1 << i;
return mask;
}
int main (int argc, char const *argv[])
{
printf("6: 0x%x\n17: 0x%x\n", mask_for_n_bits(6), mask_for_n_bits(17));
return 0;
}
0x2F is 0010 1111 in binary - this should be 0x3f, which is 0011 1111 in binary and which has the 6 least-significant bits set.
Similarly, 0x1FFFF is 0001 1111 1111 1111 1111 in binary, which has the 17 least-significant bits set.
A "mask" is a value that is intended to be combined with another value using a bitwise operator like &, | or ^ to individually set, unset, flip or leave unchanged the bits in that other value.
For example, if you combine the mask 0x2F with some value n using the & operator, the result will have zeroes in all but the 6 least significant bits, and those 6 bits will be copied unchanged from the value n.
In the case of an & mask, a binary 0 in the mask means "unconditionally set the result bit to 0" and a 1 means "set the result bit to the input value bit". For an | mask, an 0 in the mask sets the result bit to the input bit and a 1 unconditionally sets the result bit to 1, and for an ^ mask, an 0 sets the result bit to the input bit and a 1 sets the result bit to the complement of the input bit.
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.