ieee-754

问题 I'm using the following function to approximate the derivative of a function at a point: def prime_x(f, x, h): if not f(x+h) == f(x) and not h == 0.0: return (f(x+h) - f(x)) / h else: raise PrecisionError As a test I'm passing f as fx and x as 3.0. Where fx is: def fx(x): import math return math.exp(x)*math.sin(x) Which has exp(x)*(sin(x)+cos(x)) as derivative. Now, according to Google and to my calculator exp(3)*(sin(3)+cos(3)) = -17.050059 . So far so good. But when I decided to test the

问题 My SSE-FPU generates the following NaNs: When I do a any basic dual operation like ADDSD, SUBSD, MULSD or DIVSD and one of both operands is a NaN, the result has the sign of the NaN-operand and the lower 51 bits of the mantissa of the result is loaded with the lower 51 bits of the mantissa of the NaN-operand. When both operations are NaN, the result is loaded with the sign of the destination-register and the lower 51 bits of the result-mantissa is loaded with the lower 51 bits of the

问题 I'm looking to for a reasonably efficient way of determining if a floating point value ( double ) can be exactly represented by an integer data type ( long , 64 bit). My initial thought was to check the exponent to see if it was 0 (or more precisely 127 ). But that won't work because 2.0 would be e=1 m=1... So basically, I am stuck. I have a feeling that I can do this with bit masks, but I'm just not getting my head around how to do that at this point. So how can I check to see if a double is

问题 Why this code 7.30 - 7.20 in ruby returns 0.0999999999999996 , not 0.10 ? But if i'll write 7.30 - 7.16 , for example, everything will be ok, i'll get 0.14 . What the problem, and how can i solve it? 回答1: What Every Computer Scientist Should Know About Floating-Point Arithmetic 回答2: The problem is that some numbers we can easily write in decimal don't have an exact representation in the particular floating point format implemented by current hardware. A casual way of stating this is that all

问题 How can I change the 32 bit hex-value to a floating point value according to the IEEE 754? EDIT: ... data = fread(fid,1,'float32'); disp(data); ... I get this answer: 4.2950e+009 1.6274e+009 ... But how do I get 32 bit floating point (IEEE 754) numbers? 回答1: Based on one of your comments it appears that your hexadecimal values are stored as strings of characters in a file. You first want to read these characters from the file in groups of 8. Depending on the specific format of your file (e.g.

问题 I'm working on porting the sqrt function (for 64-bit doubles) from fdlibm to a model-checker tool I'm using at the moment (cbmc). As part of my doings, I read a lot about the ieee-754 standard, but I think I didn't understand the guarantees of precision for the basic operations (incl. sqrt). Testing my port of fdlibm's sqrt, I got the following calculation with sqrt on a 64-bit double: sqrt

问题 I'm working on porting the sqrt function (for 64-bit doubles) from fdlibm to a model-checker tool I'm using at the moment (cbmc). As part of my doings, I read a lot about the ieee-754 standard, but I think I didn't understand the guarantees of precision for the basic operations (incl. sqrt). Testing my port of fdlibm's sqrt, I got the following calculation with sqrt on a 64-bit double: sqrt

问题 I was reading about floating-point NaN values in the Java Language Specification (I'm boring). A 32-bit float has this bit format: seee eeee emmm mmmm mmmm mmmm mmmm mmmm s is the sign bit, e are the exponent bits, and m are the mantissa bits. A NaN value is encoded as an exponent of all 1s, and the mantissa bits are not all 0 (which would be +/- infinity). This means that there are lots of different possible NaN values (having different s and m bit values). On this, JLS §4.2.3 says: IEEE 754

问题 I am confused on IEEE754 double precision, I consider two questions: 1. Why each number from interval -2 54 , -2 54 +2, -2 54 +4...2 54 is representable ? 2. Why 2 54 +2 is not representable ? Can you help me ? I understand way of working IEEE754 - however, I have a problem with seeing it. 回答1: There are 53 bits in the significand (or mantissa) of an IEEE 754 double. −2 54 can be exactly represented, as mantissa: 1.00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00 (bin) exponent:

问题 I am confused on IEEE754 double precision, I consider two questions: 1. Why each number from interval -2 54 , -2 54 +2, -2 54 +4...2 54 is representable ? 2. Why 2 54 +2 is not representable ? Can you help me ? I understand way of working IEEE754 - however, I have a problem with seeing it. 回答1: There are 53 bits in the significand (or mantissa) of an IEEE 754 double. −2 54 can be exactly represented, as mantissa: 1.00000 00000 00000 00000 00000 00000 00000 00000 00000 00000 00 (bin) exponent: