double

问题 Comparing those two values shall result in a "true": 53.9173333333333 53.9173 回答1: If you want a = 1.00001 and b = 0.99999 be identified as equal: return Math.abs(a - b) < 1e-4; Otherwise, if you want a = 1.00010 and b = 1.00019 be identified as equal, and both a and b are positive and not huge: return Math.floor(a * 10000) == Math.floor(b * 10000); // compare by == is fine here because both sides are integral values. // double can represent integral values below 2**53 exactly. Otherwise, use

问题 Note: Similar to Can an integer be NaN in C++? I understand this has little practical purpose, but can a float or double be set to NaN ? 回答1: The Float object contains a static value, which is a float type, called NaN . So float myFloat = Float.NaN; gives you what you are asking. http://download.oracle.com/javase/6/docs/api/java/lang/Float.html#NaN 回答2: Sure! NaN is a static constant in the Float and Double classes. double x = Double.NaN; 回答3: Yes float f = Float.NaN; See the doc for more

问题 Say I have a method returning a double , but I want to determine the precision after the dot of the value to be returned. I don't know the value of the double varaible. Example: double i = 3.365737; return i; I want the return value to be with precision of 3 number after the dot Meaning: the return value is 3.365 . Another example: double i = 4644.322345; return i; I want the return value to be: 4644.322 回答1: What you want is truncation of decimal digits after a certain digit. You can easily

问题 I'm doing an integration program with Riemann sums for my Calculus class. I've decided to use C when computing my integrals, and I noticed a huge error in my program that derives from this problem. #include <stdio.h> #include <stdlib.h> #include <math.h> int main(int argc, char** argv) { double x = 2.0/20.0; printf("%1.50f \n", x); return (EXIT_SUCCESS); } The program gives me : 0.10000000000000000555111512312578270211815834045410. My question: Why does this happen? And how can I fix this? Or

问题 I tried following: std::cout << std::hex << 17.0625; But it dumped it in decimal. I'd like to see 11.01 (17.0625 in hex). How can I print some floating point value in hex? Please do not offer solutions like: void outhexdigits(std::ostream& out, fp_t d, int max_chars=160) { while(d > 0. && max_chars) { while(d < 1. && max_chars){ out << '0'; --max_chars; d*=16; } if (d>=1. && max_chars) { int i = 0; while (d>=1.) ++i, --d; out << std::hex << i; --max_chars; } } } Is there any way to dump float

问题 So in the javascript portion of my code, here is the snippet that actually sends an array of pixels to the vertex and fragment shaders- but I am only working with 1 texture when I get to those shaders- is there anyway that I can send two textures at a time? if so, how would I 'catch' both of them on the GLSL side of the codee? if (it > 0){ gl.activeTexture(gl.TEXTURE1); gl.bindTexture(gl.TEXTURE_2D, texture); gl.activeTexture(gl.TEXTURE0); gl.bindFramebuffer(gl.FRAMEBUFFER, FBO2);} else{ gl

问题 Is something broken or I fail to understand what is happening? static String getRealBinary(double val) { long tmp = Double.doubleToLongBits(val); StringBuilder sb = new StringBuilder(); for (long n = 64; --n > 0; tmp >>= 1) if ((tmp & 1) == 0) sb.insert(0, ('0')); else sb.insert(0, ('1')); sb.insert(0, '[').insert(2, "] [").insert(16, "] [").append(']'); return sb.toString(); } public static void main(String[] argv) { for (int j = 3; --j >= 0;) { double d = j; for (int i = 3; --i >= 0;) { d +

问题 This question already has answers here : Is floating point math broken? (31 answers) Closed 3 years ago . For some reason, certain Doubles in my Swift app are giving me trouble when converting to NSNumber, while some are not. My app needs to convert doubles with 2 decimal places (prices) to NSNumbers so they can be stored and retrieved using Core Data. For example, a few particular prices such as 79.99 would evaluate to 99.98999999999999 unless specifically formatted using NSNumber's

问题 All began with these simple lines of code: string s = "16.9"; double d = Convert.ToDouble(s); d*=100; The result should be 1690.0, but it's not. d is equal to 1689.9999999999998. All I want to do is to round a double to value with 2 digit after decimal separator. Here is my function. private double RoundFloat(double Value) { float sign = (Value < 0) ? -0.01f : 0.01f; if (Math.Abs(Value) < 0.00001) Value = 0; string SVal = Value.ToString(); string DecimalSeparator = System.Globalization

问题 What I'm trying to do is to convert a double to hex string and then back to double. The following code does conversion double-to-hex string. char * double2HexString(double a) { char *buf = new char[17]; // double is 8-byte long, so we have 2*8 + terminating \0 char *d2c; d2c = (char *) &a; char *n = buf; int i; for(i = 0; i < 8; i++) { sprintf(n, "%02X", *d2c++); n += 2; } *(n) = '\0'; } This seems work, however, I'm not sure how to convert the resulting string back to double. Please advise :