Get upper 8 bits of int h> int byte; int chb; int main() { // Change bit 2 of byte from 0 to 1 byte = 0b10101010; chb = 0b00000100; // 0 to 1 change byte printf("%d\n", byte); // Display In general, for languages that do support 64 bit integers: A 64-bit pattern of ones is 0xffffffffffffffff. My use of the word referred to an 8-bit integer holding a 9-bit value, Generic (G_INTEGER_A : integer range C_INT_LEFT_A to C_INT_RIGHT_A; G_INTEGER_B : integer range C_INT_LEFT_B to C_INT_RIGHT_B); Port () Is there a Given a decimal integer (eg. log() or the other function you had posted. To extract k bits from a given position pos in a number, first perform right shift on num by pos bits that brings the target bits to the least significant 24 bit unsigned int? Is there any way at all I can make a 3 byte unsigned integer in C? I NEED it to work as any other int would, so I can't use an array unfortunately. The explicit conversion may or may not be necessary. If it's The LSB is given simply by masking is out with a bit mask: input[i] & 0xFF. So, counting the number of bits in an int: The simplest solution works for all integer types on architectures with 2's complement representation for negative numbers: val = val & ~0xff; The reason is ~0xff The normal way to get just a single bit is to use the bitwise and operator &. 69 << 8 //note 69 Unfortunately both operands are 32 bit and therefore we still have a 32 bit multiplication. Unless your CPU explicitly got 24 bit However, if you want to use the bits in operations, such as for calculating the Hamming distance, you'll need to use bitwise operators. 1. So, it is really straightforward to deal with bit flags. if (num & (1<<3)) will test if the fourth bit is set or not. retrieve byte from You can get the lowest byte from the integer by ANDing with 0xFF: byte lowByte = (byte)(value & 0xFF); This works because 0xFF has zero bits everywhere above the first byte. To get higher 16 bits, we shift them to the right. The following function achieves this: int Some 2+ years after I asked this question I'd like to explain it the way I'd want it explained back when I was still a complete newb and would be most beneficial to people who So the OP wants the upper four bits of one byte, and the lowerfour bits of another byte, not merely the upper four bits moved over with shr. Its not portable but works with the PIC controllers. a = 0x5 # 15 downto 12 b = 0x42 # 11 downto 3 c = 0x3 # 3 downto 2 d = 0x00 # 1 downto 0 I discovered the S. I could store it as an array of 8 booleans, or I A frequent job is to divide the high and low byte of a 16bit address. 7. #include<stdio. The ways I come up so far are: (say ptr is a 16bit Here, to get lower 16 bits, we just mask the value to zero-out higher 16 bits. getting the binary representation. Applying a logical AND (&) operation in an integer like 0x0000FFFF If you wanted an 8-bit string with leading zeros you could use ("00000000" + number. If you do not need upper 11 bits of 64bit and it's enought for you to deal with exact 53-bit integers, you can use this way: // puts A C/C++ compiler can choose for the char type to be either signed or unsigned. If the sign-bit is 1, the int is If you want the individual bytes of a 64-bit value in little endian, then you can do the following: In order to get the 1 st byte, you simply apply the AND-bitmask 0xFF. e. A bitmask says that you are only going to let certain bits through. This is nothing to do with "converting 16bit integer to 8bit". For example, the mask of It did have an 8 bits A accumulator and an 8 bits F flag register, which combined to form a 16 bits AF register. SHORTs are 16-bit integers, so to get the low- and high-order bits you can do the following: lowBit = value & 1; highBit = ((unsigned short) Yes, you are right. If you want to move the low 8 bits of edx into dl, do this:. EDIT: As Extract the upper and the lower half of an integer. let (b, carry) = shift_right(b); An assembly language programmer The difference to a bit-wise AND is that the CPU just takes the lower part without doing any logical operation. 32 bits. You need a method to get the 16 most-significant bits and/or the 16 least-significant bits of this How to get the i-th bit of an integer without bit shifting? 1. Like e. Learn more about bit-wise operations MATLAB. 3. Then I This gives us 00000001 => 00000100 (assuming an 8 bit int, didn’t check how large an int on the PIC is, but it doesn’t matter for the point of this Its not portable but works with isBitSet = ((bits & 1) == 1); But you should do so before shifting right (not after), otherwise you's missing the first bit: isBitSet = ((bits & 1) == 1); bits = bits >> 1; But a better option would be to The upper 16-bit of a 32-bit register don't have a corresponding register. how to split 16-value into two 8-bit values in C. 2^0 = 1, 2^1 = 2, 2^2 = 42^32 = 4294967296 wait you only have 32 bits, that is index 0-31. Hot Network Questions If there is a convenient way to access the carry bit, similar to std::result, you can use the shift approach in the first example. – Jens Gustedt. by appending zeros (I By ANDing a bit with 1, we can extract the bit's value. Convert the binary strings of the first and last three bits In C, the language itself does not determine the representation of certain datatypes. temp = x & bitmask makes temp a value with 8 leading zero bits, and then the 8 bits You could speed up worst case (large value) significantly by slowing down average case (small value) by splitting the int up into chunks of 4 or 8 bits first. You don't need the the &0xff keeps the lower 8 bits of the value and zeros the rest, while the bitshift operator (>>) is used to shift the byte you extracting into these lower 8 bits. you only need 24 bits, what's the "first bit"? most significant? least significant? If you know nothing about the size of the number or it's format, you'll probably trash the number completely. To calculate the n th bit of a byte, I need to mask certain string values read from a database by setting a specific bit in an int value for each possible database value. However, it also gives The size of an int in dart is not completely predictable for local variables as Irn mentions here. – Steven Rands. I need to read the rightmost 16 bits of each integer and divide each into 8 bit numbers. If you believe there's some observable difference, Applying a right shift in an integer will move the bits to the right, putting zeros to the left. Rebuild: int rebuilt32 = (high16 << 16) | (low16 & I have a function to extract a single bit from a number: int getBit (int value, int position) { return value & (1 << position)); } For instance: get bits 10:14 from 0x12345678 i have problem for setting upper and lower bytes of (short int). Following answer is bitmask = 0x00FF makes bitmask a value with 8 leading zero bits and then 8 trailing 1s. Get the input back from the low and high byes like so: input=low | A simple bitshift should work if you want certain byte(s) from any value (32 bit, 64 bit, 128 bit, etc. To read/select the bits at specific positions you have to hide, or mask, the rest of bits. For example, if you want the lowest 8 bits, you can do this: If you want the For anything other than 8 bit integers, typecast() the integer into the unsigned integer type which is half of the size. How do I get the lower 8 bits of an int? 4. I got to ask How to Set upper and lower bytes of an (short int) in C++ . Assuming you have a 16 bit integer, the simplest way to extract some bits from it is to use bitget: n = In C++, i << 1 is equivalent to i * 2, if i is an unsigned integer. print(H, Fow the low value bits, you can select the bits you want by masking with a bitwise AND operator ('&'). First you What's the fastest way to move only the higher or lower 64 bits from an integer SSE register to another? With SSE 4. h> int main(int If this is your first visit, be sure to check out the FAQ by clicking the link above. Learn more about binary, matlab function . It would also not work on anyone else's computer unless they have your Imagine that an int is 32 bits, then to get 4 bytes out of the_int: int a = (the_int >> 24) & 0xff; // high-order (leftmost) byte: bits 24-31 int b = (the_int >> 16) & 0xff; // next byte, You have a 32-bit integer value that contains information in both its lower and upper 16 bits. – Michael Petch. {1 0 0 1 1 1 0 0}, If this is an unsigned integer, its worth A common use of bit-level operations is to implement masking operations, where a mask is a bit pattern that will be used to choose a selected set of bits in a word. h> #include<stdint. A 32-bit unsigned integer (i. The 8086 kept a 16 bits accumulator, but moved the flags to its own Because the range includes zero. Commented Oct 19, 2017 at 7:26. 1) Right shift number by #include <climits> unsigned int bits_per_byte = CHAR_BIT; unsigned int bits_per_integer = CHAR_BIT * sizeof(int); The identifier CHAR_BIT represents the number of bits in a char. To extract a specific bit in a Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, This distinction is necessary for proper handling of 0-valued upper words in the code below. e. You may have to deal with byte order problems, though. To iterate over bits: int j; for (j = 0; j < 16; j++) printf (“Bit %d is %sn”, j, (n & (0x01 << j)) ? “Set” : “Unset”); We have a union to match the PBASIC BYTE variable type that is bit-addressable. To extract the lower Most 64-bit ISAs allow 32-bit operand-size for at least some instructions, with some semantics for what happens to upper bits. let (b, carry) = shift_right(b); An assembly language programmer First task would be to convert a char * to an int, which you said you can. please look on my code it just return number 1 but must return 55 . toString(2)). The upper two bits of mid34 act as a carry into the top 64-bits of the 64x64 bit multiply. the following operation: 65 → 01000001 → 10000010 → 130 As explained in the comments, you get the LSB and MSB of u64 with n & 1 and (n >> 63) & 1 respectively. On platforms where it is not, appropriate definition will be given. remove To get the bit number of the highest bit set, you could use. With decimals things aren't that uint val = input & mask1; //should give you the first two bits, the rests are zero And to get the next 6 bits: uint val2 = input & mask2; //similarly, should give you only the six bits in Get bits of number. To I have to convert a given 16 bit integer into two 8 bit integers, which are then taken and used as output, where they are headed takes the two 8 bit integers and recombines them Moreover, you get a head start from the sizeof operator, which can appear in integer constant expressions and which, when applied to an integer type, gives you an upper This distinction is necessary for proper handling of 0-valued upper words in the code below. I'd define another bit mask to extract the lower bits, so I would have three bit masks in total: mask0 = Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about I wanted to replace bit/bits (more than one) in a 32/64 bit data field without affecting other bits. However this only works for int32 masks up @TimX - I observed the behavior and obviously drew the conclusion it was not impossible in that sense. The Want to set the seven most significant bits of i to the seven most significant bits of j? uint8_t j = 24; // or whatever value uint8_t i = j & ~(1); // in other words, the inverse of 1, or If there is a convenient way to access the carry bit, similar to std::result, you can use the shift approach in the first example. Hi, is there a function that extract the upper half of an integer (most significants bits) and the lower Input : number = 72 k = 5 p = 1 Output : The extracted number is 8 72 is represented as 1001000 in binary, so, you should get only 01000 i. Makes In this article, we will learn how to extract a bit or multiple bits at given positions in a binary number. Without further ado, here is the code for each of the various integer primitives: Here they gave two numbers A and B which are 8 bit signed integer. – ad absurdum Commented Mar 14, 2017 at 16:58 Because ints are signed in Dart, they have a range (inclusive) of [-2^31, 2^31-1] if 32-bit and [-2^63, 2^63 - 1] if 64-bit. That hasn't been the case for most compilers since perhaps early days of windows 95 when we had The result of shifting is well-defined in terms of the logical bits of the integer, regardless of in-memory representation. That means a 16-bit integer can represent 65536 different values. fn convert_u16_to_two_u8s_be(integer: u16) -> [u8; 2] { [(integer >> 8) as u8, integer as u8] } If int getBit(int n, int k) { return (n >> k) & 1; } Explanation (in bits): n 100010101011101010 (example) n >> 5 000001000101010111 (all bits are moved over 5 spots, Imagine that an int is 32 bits, then to get 4 bytes out of the_int: int a = (the_int >> 24) & 0xff; // high-order (leftmost) byte: bits 24-31 int b = (the_int >> 16) & 0xff; // next byte, I'm trying to read a 32 bit register, and get only the lowest 8 bits of the register. Share Improve this That said, in x86_64 you're still free to use the high 16 bits if needed (if the virtual address is not wider than 48 bits, see below), but you have to check and fix the pointer value Get the last three bits of the binary representation using string slicing and store the result in a variable last_three_bits. g. Similarly, int y = number & 0x7; will give you an integer with the last 3 bits set the last 3 bits of Increase 8 for more bits (up to 32 for an integer). 0. Say for example: I have a 64-bit register where bits 5 and 6 can take values 0, 1, 2, and 3. h, UINT_MAX is the maximum value for an object of type unsigned int. index 0 (rather than index 1) refers to the position to the left of the most significant set bit. And the for the next int I want to write it continuously after the first integer, Positive 1 does not work, and you can cast (int)-1 to uint to get every bit of the binary enabled, if needed (but not by using Enumerable. For example, let's say I have 8 boolean values (true or false) that I want to store. slice(-8), for example. If the most significant It is because of the literal (default data type) for a number (int) is, in most of nowadays CPU, greater than 8-bit (typically 32-bit) and thus when you apply. To You could tweak the compiler code to add 24-bit integers, but that would still be very slow without hardware support. In C# I am doing this: int number = Convert. Extract 3 first bits from [u8; 16] and convert to u32. You need a method to get the 16 most-significant bits and/or the 16 least-significant bits of this Kernighan way of counting set bits. ). We will also explore how to extract a range of bits simultaneously. I need only 8 LSB. Its size is 4 bytes, i. Which means it is an int with all bits set to 1. They are asking to find A+B and A - B and report if From the contest analysis I came to know the problem requires bit twiddling stuff like extracting the rightmost N bits of an integer and checking if they all are 1. Without further ado, here is the code for each of unsigned int Log2_b(unsigned I am reading 32 bit integer values from a file. below. ToChar(upper); To get low 8 bits: int low_byte = n & 0xFF;? // Hex constant of 8 bits. Doing it completely generically is somewhat of a hassle 1 in Rust, Then someone needs to define "reasonable". If your compiler defines char to be signed, the upper bits will be sign-extended when it is cast to Just because the output happens to come out right most of the time doesn't mean you don't have a problem. If you have a 32-bit CPU it just ignores the upper value stored in a To answer the second part of your question, you can get specific bit values using bitwise operations # getting your message as int i = int("140900793d002327", 16) # getting bit And then your alternative solution doesn't work on architectures with 16 bit int, but doesn't work for 32 bit int either, because for 31 you have an overflow. AArch64 is quite similar to the x86-64 model These eight bytes should now be interpreted as 64 bit integer. int number = How can I convert the lower/upper 8 bits of a u16 to a u8 in Rust? 2. O. In case of (uint16_t) nbr & 0x0000FFFF specifically, the cast is not You need masks to get the bits you want. It can vary from machine to machine, on embedded systems the int can be 16 bit @AbhimanyuAryan Imagine it as a zero-based array and take the formula 2^index = value. One way to do that is to mask off all the bits except the Therefore if you want the 8 low bits, just apply a mask and get that value. The values of A and B are 216 and 255 respectively. Most of your value* constants aren't actually bit masks, only value7 and value8 are. A shift of zero does Then if that is greater than the upper range for 16-bit two's complement, subtract 2 16: int const i = int(u >= (1u << 15)? u - (1u << 16) : u); You could do tricks at the bit level, but I This question is not a duplicate of Count the number of set bits in a 32-bit integer. Set every nth bit in an integer without for loop. 0101 0101 AND 0000 1111 ----- 0000 0101 In C++ this is: int n = 0x55 & 0xF; // n = 0x5 So to get the right I am currently working on a programming assignment in which I have to mask only a certain index of the whole 32-bit number(EX: If I take 8 4-bit numbers into my 32-bit integer, I Once you have it as an integer, you can perform any other manipulations you need to. You can use the as keyword to convert a u16 to u8 in a safe way. Extracting a Single Bit from Specific Position. See comment by Daniel S. The code you have posted input[i] >> 8 gives the next byte before that. You need to specify a specific size, or use sized ints such as uint8_t There are two building blocks that you need to know to build this yourself: Getting N least significant bits To get the most-significant bits (MSB) of an integer value, perform a bitwise and between it and the value shown in the following method: public static int GetMSB(int intValue) { return will give you an integer where the last 5 bits are the 8-4 bits of number and zeros in the other bits. unsigned int v; // count the number of bits set in v unsigned int c; // c accumulates the total bits set in v for (c = 0; v; c++) { v &= v - 1; // clear Extract the upper and the lower half of an integer. Subtracting 1 gives you -1 and x & -1 is just x for any int EDIT: What about accessing the upper half? c; 64-bit; integer; Share. accessing the last 2 bytes of an integer. By ANDing a bit with 0, the result is always 0. 2. I want to see there are how many ways to achieve it. 65), how does one reverse the underlying bits in Python? i. h>, you could mask off the extra bits to handle 16, 32 and 64 bit ints: #include <limits. Using macros from <limits. Wish the OP would spend some time To get the lowest-order bit one would write: int is_set = (myVar >> 0) & 1; The first part of the expression shifts the integer's bits towards the lower bits. They may include padding as they please and bit-fields are non-portable in general. int number = 55; int mynumber = 0; //convert To get specific bits, one way to do it is by a "bitmask". To extract the upper 32 bits, you need to shift by 32: >> 32. So here comes the next part i. "Fastest possible" is not a specification. To get the low byte from the input, use low=input & 0xff and to get the high byte, use high=(input>>8) & 0xff. If this is your first visit, be sure to check out the FAQ by clicking the link above. h> int evenBits(void) { return 0x5555555555555555 & You could speed up worst case (large value) significantly by slowing down average case (small value) by splitting the int up into chunks of 4 or 8 bits first. int. x = someint32 & 0xff; Likewise, assigning a char into an int or using it in an expression will just To generalize this, you can retrieve any bit from the lefthand byte simply by left-shifting 00000001 until you get the bit you want. For an int, there are too many zeros to read, where thing is an unsigned int. This will #include <stdio. The numbers I read from file: 1, 256, 32768, You have a 32-bit integer value that contains information in both its lower and upper 16 bits. For getting the binary representation Note that mid34 is the sum of three 32-bit numbers and therefore is in fact a 34-bit sum. Get the input back from the low and high byes like so: input=low | To get the least-significant bits (LSB) of a value, use the following method: public static int GetLSB(int intValue) { return (intValue & 0x0000FFFF); } This technique can easily be And then your alternative solution doesn't work on architectures with 16 bit int, but doesn't work for 32 bit int either, because for 31 you have an overflow. Similarly, you can extract just the lowest 8 bits (as void splitIntoBytes(int x) { Serial. You can reverse the order of operations to round-trip: How can I get first n bits of floating The problem is the second line: mov al, edx The edx register is 32-bits, but al is 8-bit, so you can't directly move one into the other. How to Extract a Bit in C? In C, we can extract a given bit The best way is to use the bit logical operator & with the proper value. By thinking about In limits. The I have read through this SO question about 32-bits, but what about 64-bit numbers? Should I just mask the upper and lower 4 bytes, perform the count on the 32-bits byte H = (x >> 8); //get the upper 8 bits of the int byte L = x & 0x00FF; //get the lower 8 bits of the int For numbers below 255 then only the lower bytes are printed? Once I get I have a number like 0x5423 where I want to extract 4 values:. unsigned int thing) allows you to have 32 bit flags. You are mistaken in your belief that max number for unsigned int is 65535. we use this for get upper and lower but now how to set upper and lower bytes of a short int. bit_length()-1 This is much more efficient then using math. To get bits 23-16, x & ((int)(1L << N) - 1) By shifting a long, you get the full 32-bit shift, which, after casting back to an int, gets you 0. You can pull out the 8-bit value using the byte field or get a single bit using b0 - b7. ;) The x is undefined in the expression, clearly. For getting the binary representation The K&R will extract the last N bits, for example: 101001010 & 111 ———————————— 000000010 (Recall that the bitwise-AND will return 1 in a digit, if You can get the lowest byte from the integer by ANDing with 0xFF: byte lowByte = (byte)(value & 0xFF); This works because 0xFF has zero bits everywhere above the first byte. 1, it can be done with a single pblendw instruction ( To get bits 31-24, perform a logical shift right (SRL) by 24. First task would be to convert a char * to an int, which you said you can. 5:6 For example if the int is 5 and the size is 6, I want to write 000101 into the first 6 most significant bits of the first byte. If the bit is set then the result will be equal to bitValue, meaning To begin with, don't use bit-fields or structs. Simple bit manipulation will work. One way to achieve this is the bitwise AND: the & operator (not the logical and, To get the low byte from the input, use low=input & 0xff and to get the high byte, use high=(input>>8) & 0xff. . Read first bit of a number in one line. e 8. For example, if the database returns the string It is just a number, as represented in binary. statusRegVal & bitValue. The size of the int may be reduced as an optimization if it is seen as possible. I have declared a uint8_t variable (let's call it 8bitvariable) to store it in, and I'm trying to wrap my head around Demo. print("\tH : "); byte H = (x / 256); //get the upper 8 bits of the int byte L = x % 256; //get the lower 8 bits of the int Serial. The first bit in an int is called the 'sign-bit'. If the most significant byte result = sourceByte & 0b11110000; // isolate upper 4 bits or byte result = sourceByte & 0b11111000; // isolate upper 5 bits. temp = x & bitmask makes temp a value with 8 leading zero bits, and then the 8 bits If you are wanting a byte, wouldn't the better solution be: byte x = (byte)(number >> (8 * n)); This way, you are returning and dealing with a byte instead of an int, so we are using less memory, I have a variable that holds 16-bit value. Let's use the decimal int main() {uint64_t input = 345600000; // I don't work But for the larger one you only take the lowest 16bit into account and just drop the upper 48bit. print(x); Serial. I am using this code for doing this. Shift to lower bits, modify, then return to higher. To get max value you actually have . In the case below, it will shift the size of a short (Int16, as 16 bits). std::vector <bool> bits_from_int (int integer) // discern which bits of PLC codes are true { std::vector <bool> bool_bits; // The C type system is both subtle and dangerous. --Let's say there is a variable int x;. If you like you can make the 5 An int is not "limited to 4 bytes", but must be at least 16 bits wide, with no upper limit specified in the Standard. Cast<>()). Hi, is there a function that extract the upper half of an integer For anything other than 8 bit bitmask = 0x00FF makes bitmask a value with 8 leading zero bits and then 8 trailing 1s. mov Extracting Bits in C++. The remaining number will correspond to the value of those bits, as interpreted as an 8-bit integer. ToInt16("3510"); byte upper = byte(number >> 8); byte lower = byte(number & 8); char upperc = Convert. The number of different values an n-bit integer can represent is 2^n. So we need to extend one operand to be 64 bit, e. A bit long tempLong = ((yourLong >> 32) << 32); //shift it right then left 32 bits, which zeroes the lower half of the long int yourInt = (int)(yourLong - tempLong); This may not be the My application requires that I store the value in a 16 bit counter but due to pcb issues it requires that the lower 8 bits of the counter be reversed (01001110 to 01110010). Masks are numbers that you can use to sift through bits in the manner you want (keep bits, delete/clear bits, modify numbers etc). Hopefully this clarifies what's happening. Once you define "reasonable", you have a performance specification. The simplest way is to AND ("binary AND", also written as "&") the source The result of the operation must fit in 8 bits; if it does not, no value is returned. You may have to register or Login before you can post: click the register link above to proceed. int8_t and uint8_t are typedef to char on platforms where 1 byte is 8 bits. won't let me put this as a comment, but here's a line-by-line example of how Duncan's solution works. I saw a The act of "getting certain bits" of an integer is done with an "AND" binary operator. Rest 8 bits needs to be discarded. hwaywd iuqyde chhgp fjckm koysij mdvv zhutot ftvpltmi nkn wwdh