ieee-754

问题 What is the risk of precision loss associated with maths like the following when using IEEE754 floating point numbers (in JavaScript)? 10*.1 i.e. integers multiplied by a rational number. 回答1: Note: The question was edited to add "that is a divisor of" long after this answer was posted, see below the fold for an update. What is the risk of precision loss... It's virtually guaranteed , depending on the integer and floating point number involved, because of the mismatch between what we use

问题 I have been researching about portable way to store a float in a binary format (in uint64_t), so that it can be shared over network to various microcontroller. It should be independent of float's memory layout and endianness of the system. I came across this answer. However, I am unable to understand few lines in the code which are shown below: while(fnorm >= 2.0) { fnorm /= 2.0; shift++; } while(fnorm < 1.0) { fnorm *= 2.0; shift--; } fnorm = fnorm - 1.0; // calculate the binary form (non

问题 I am getting 4 bytes of data through an interface(Bluetooth, List). The data is representing IEEE 754 float (e.g. 0x3fd0a3d7 , which represents approximately 1.63 as a binary32 float ) Is there a way in dart lang to convert / type-pun this to float and then double? Something like intBitsToFloat in Java. Couldn't find anything. Or do I just have to write the IEEE 754 parsing myself? 回答1: This is working, just import the dart:typed_data library: var bdata = ByteData(4); bdata.setInt32(0,

问题 This really shouldn't be this difficult. I want to read raw 64-bit IEEE 754 double-precision floating-point data from a file, and use it in a std_logic_vector(63 downto 0) . I'm using ModelSim ALTERA 10.1b. I tried to just read the raw binary data into the 64-bit vector: type double_file is file of std_logic_vector(63 downto 0); file infile1: double_file open read_mode is "input1.bin"; variable input1 : std_logic_vector(63 downto 0) := (others => '0'); read(infile1, input1); But this doesn't

问题 My impression is that C float has 8 bits of exponent and 23 bits of mantissa. So one is 0011 1111 1000 0000 0000 0000 0000 0000 = 0x3F800000. However, the following code produced 1.06535e+09 instead of 1. Can anyone help me understand why? #include <iostream> #include <math.h> using namespace std; int main() { float i = 0x3F800000; cout<<i << endl; return 0; } 回答1: How is 1 coded in C as a float? Can anyone help me understand why (code fails)? float i = 0x3F800000; is the same as i =

问题 I assumed that the minimal positive value that can be put into double floating point is this 0 0000000000 00000000000000000000000000000000000000000000000000000 And in scientific form it's this: 1 x 2^{-1023} However, this article states that: As mentioned above, zero is not directly representable in the straight format, due to the assumption of a leading 1 (we'd need to specify a true zero mantissa to yield a value of zero). Zero is a special value denoted with an exponent field of all zero

问题 I am trying to come up with an efficient method to determine when rounding will/did occur for IEEE-754 operations. Unfortunately I am not able to simply check hardware flags. It would have to run on a few different platforms. One of the approaches I thought of was to perform the operation in different rounding modes to compare the results. Example for addition: double result = operand1 + operand2; // save rounding mode int savedMode = fegetround(); fesetround(FE_UPWARD); double upResult =

问题 I am trying to come up with an efficient method to determine when rounding will/did occur for IEEE-754 operations. Unfortunately I am not able to simply check hardware flags. It would have to run on a few different platforms. One of the approaches I thought of was to perform the operation in different rounding modes to compare the results. Example for addition: double result = operand1 + operand2; // save rounding mode int savedMode = fegetround(); fesetround(FE_UPWARD); double upResult =

问题 Entering a value such as 27.8675309 into the "Decimal representation" field of the IEEE 754 Converter changes the value I entered to 27.86753 . Likewise, Java drops the last two digits when a parse a string with the same value. Float.parseFloat("27.8675309") // Results in a float value of 27.86753 I am not sure what the "Decimal representation" of the IEEE converter actually is (is it a float?) but I would expect it to give me the biggest number possible that: Is a float value Does not exceed

问题 I need to do IBM-IEEE floating point conversions of byte arrays in Java. I was able to successfully do conversions of single-precision float bytes using http://www.thecodingforums.com/threads/c-code-for-converting-ibm-370-floating-point-to-ieee-754.438469. But I also need to convert bytes of double-precision doubles. Everywhere I look seems to only show the single-precision conversion. The closest I've gotten is the r8ibmieee function in http://spdf.sci.gsfc.nasa.gov/pub/documents/old