Search Results

Search found 950 results on 38 pages for 'floating'.

Page 3/38 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Using write to print floating point numbers.

    - by Tom
    Hi, As an exercise to achieve something larger, i'm trying to use write to print a floating point number. I haven't done this in a while. I must be doing something wrong because I cant get it to work. Here is my code #include <unistd.h> int main(){ float f = 4.5; write(1,&f,sizeof float); return 0; } However, when running it im getting ?@ Any thoughts? Thanks in advance.

    Read the article

  • Floating point innacuracies

    - by Greg
    While writing a function which will perform some operation with each number in a range I ran into some problems with floating point inaccuracies. The problem can be seen in the code below: #include <iostream> using namespace std; int main() { double start = .99999, end = 1.00001, inc = .000001; int steps = (end - start) / inc; for(int i = 0; i <= steps; ++i) { cout << (start + (inc * i)) << endl; } } The problem is that the numbers the above program outputs look like this: 0.99999 0.999991 0.999992 0.999993 0.999994 0.999995 0.999996 0.999997 0.999998 0.999999 1 1 1 1 1 1.00001 1.00001 1.00001 1.00001 1.00001 1.00001 They only appear to be correct up to the first 1. What is the proper way to solve this problem?

    Read the article

  • floating point equality in Python and in general

    - by eric.frederich
    I have a piece of code that behaves differently depending on whether I go through a dictionary to get conversion factors or whether I use them directly. The following piece of code will print 1.0 == 1.0 -> False But if you replace factors[units_from] with 10.0 and factors[units_to ] with 1.0 / 2.54 it will print 1.0 == 1.0 -> True #!/usr/bin/env python base = 'cm' factors = { 'cm' : 1.0, 'mm' : 10.0, 'm' : 0.01, 'km' : 1.0e-5, 'in' : 1.0 / 2.54, 'ft' : 1.0 / 2.54 / 12.0, 'yd' : 1.0 / 2.54 / 12.0 / 3.0, 'mile' : 1.0 / 2.54 / 12.0 / 5280, 'lightyear' : 1.0 / 2.54 / 12.0 / 5280 / 5.87849981e12, } # convert 25.4 mm to inches val = 25.4 units_from = 'mm' units_to = 'in' base_value = val / factors[units_from] ret = base_value * factors[units_to ] print ret, '==', 1.0, '->', ret == 1.0 Let me first say that I am pretty sure what is going on here. I have seen it before in C, just never in Python but since Python in implemented in C we're seeing it. I know that floating point numbers will change values going from a CPU register to cache and back. I know that comparing what should be two equal variables will return false if one of them was paged out while the other stayed resident in a register. Questions What is the best way to avoid problems like this?... In Python or in general. Am I doing something completely wrong? Side Note This is obviously part of a stripped down example but what I'm trying to do is come with with classes of length, volume, etc that can compare against other objects of the same class but with different units. Rhetorical Questions If this is a potentially dangerous problem since it makes programs behave in an undetermanistic matter, should compilers warn or error when they detect that you're checking equality of floats Should compilers support an option to replace all float equality checks with a 'close enough' function? Do compilers already do this and I just can't find the information.

    Read the article

  • Keeping sync in multiplayer RTS game that uses floating point arithmetic

    - by Calmarius
    I'm writing a 2D space RTS game in C#. Single player works. Now I want to add some multiplayer functionality. I googled for it and it seems there is only one way to have thousands of units continuously moving without a powerful net connection: send only the commands through the network while running the same simulation at every player. And now there is a problem the entire engine uses doubles everywhere. And floating point calculations are depends heavily on compiler optimalizations and cpu architecture so it is very hard to keep things syncronized. And it is not grid based at all, and have a simple phisics engine to move the space-ships (space ships have impulse and angular-momentum...). So recoding the entire stuff to use fixed point would be quite cumbersome (but probably the only solution). So I have 2 options so far: Say bye to the current code and restart from scratch using integers Make the game LAN only where there is enough bandwidth to have 8 players with thousands of units and sending the positions and orientation etc in (almost) every frame... So I looking for better opinions, (or even tips on migrating the code to fixed-point without messing everything up...)

    Read the article

  • Make floating element "maximally wide"

    - by bobobobo
    I have some floating elements on a page. What I want is the div that is floated left to be "maximally wide" so that it is as wide as it possibly can be without causing the red div ("I go at the right") to spill over onto the next line. An example is here: The width:100%; doesn't produce the desired effect! ** I don't want the green element ("I want to be as wide as possible") to go "under" the red element. Its very important that they both stay separate i.e. .. I think they must both be floated! <div class="container"> <div class="a1">i go at the right</div> <div class="a2">i want to be as wide as possible,</div> <div class="clear"></div> </div> <style> div { border: solid 2px #000; background-color: #eee; margin: 8px; padding: 8px; } div.a1 { float:right; background-color: #a00; border: solid 2px #f00; margin: 12px; padding: 6px; } div.a2 { float: left; /*width: 100%;*/ /*this doens't produce desired effect!*/ background-color: #0b0; border: solid 2px #0f0; margin: 12px; padding: 14px; } .clear { border: none; padding: 0 ; margin: 0; clear:both; } </style>

    Read the article

  • c++ floating point precision loss: 3015/0.00025298219406977296

    - by SigTerm
    The problem. Microsoft Visual C++ 2005 compiler, 32bit windows xp sp3, amd 64 x2 cpu. Code: double a = 3015.0; double b = 0.00025298219406977296; //*((unsigned __int64*)(&a)) == 0x40a78e0000000000 //*((unsigned __int64*)(&b)) == 0x3f30945640000000 double f = a/b;//3015/0.00025298219406977296; the result of calculation (i.e. "f") is 11917835.000000000 (*((unsigned __int64*)(&f)) == 0x4166bb4160000000) although it should be 11917834.814763514 (i.e. *((unsigned __int64*)(&f)) == 0x4166bb415a128aef). I.e. fractional part is lost. Unfortunately, I need fractional part to be correct. Questions: 1) Why does this happen? 2) How can I fix the problem? Additional info: 0) The result is taken directly from "watch" window (it wasn't printed, and I didn't forget to set printing precision). I also provided hex dump of floating point variable, so I'm absolutely sure about calculation result. 1) The disassembly of f = a/b is: fld qword ptr [a] fdiv qword ptr [b] fstp qword ptr [f] 2) f = 3015/0.00025298219406977296; yields correct result (f == 11917834.814763514 , *((unsigned __int64*)(&f)) == 0x4166bb415a128aef ), but it looks like in this case result is simply calculated during compile-time: fld qword ptr [__real@4166bb415a128aef (828EA0h)] fstp qword ptr [f] So, how can I fix this problem? P.S. I've found a temporary workaround (i need only fractional part of division, so I simply use f = fmod(a/b)/b at the moment), but I still would like to know how to fix this problem properly - double precision is supposed to be 16 decimal digits, so such calculation isn't supposed to cause problems.

    Read the article

  • Displaying Multimedia Content In A Floating Window Using FancyBox

    While surfing the web you may have come across websites with images and other multimedia content that, when clicked, were displayed in a floating window that hovered above the web page. Perhaps it was a page that showed a series of thumbnail images of products for sale, where clicking on a thumbnail displayed the full sized image in a floating window, dimming out the web page behind it. Have you ever wondered how this was accomplished or whether you could add such functionality to your ASP.NET website? In years past, adding such rich client-side functionality to a website required a solid understanding of JavaScript and the "eccentricities" of various web browsers. Today, thanks to powerful JavaScript libraries like jQuery, along with an active developer community creating plugins and tools that integrate with jQuery, it's possible to add snazzy client-side behaviors without being a JavaScript whiz. This article shows how to display text, images, and other multimedia content in a floating window using FancyBox, a free client-side library. You'll learn how it works, see what steps to take to get started using it, and explore a number of FancyBox demos. There's also a demo available for download that shows using FancyBox to display both text and images in a floating window in an ASP.NET website. Read on to learn more! Read More >

    Read the article

  • Displaying Multimedia Content In A Floating Window Using FancyBox

    While surfing the web you may have come across websites with images and other multimedia content that, when clicked, were displayed in a floating window that hovered above the web page. Perhaps it was a page that showed a series of thumbnail images of products for sale, where clicking on a thumbnail displayed the full sized image in a floating window, dimming out the web page behind it. Have you ever wondered how this was accomplished or whether you could add such functionality to your ASP.NET website? In years past, adding such rich client-side functionality to a website required a solid understanding of JavaScript and the "eccentricities" of various web browsers. Today, thanks to powerful JavaScript libraries like jQuery, along with an active developer community creating plugins and tools that integrate with jQuery, it's possible to add snazzy client-side behaviors without being a JavaScript whiz. This article shows how to display text, images, and other multimedia content in a floating window using FancyBox, a free client-side library. You'll learn how it works, see what steps to take to get started using it, and explore a number of FancyBox demos. There's also a demo available for download that shows using FancyBox to display both text and images in a floating window in an ASP.NET website. Read on to learn more! Read More >Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

    Read the article

  • Exact textual representation of an IEEE "double"

    - by CyberShadow
    I need to represent an IEEE 754-1985 double (64-bit) floating point number in a human-readable textual form, with the condition that the textual form can be parsed back into exactly the same (bit-wise) number. Is this possible/practical to do without just printing the raw bytes? If yes, code to do this would be much appreciated.

    Read the article

  • Why does 99.99 / 100 = 0.9998999999999999

    - by the-locster
    Whereas 99.99 * 0.01 = 0.99 Clearly this is the age old floating point rounding issue, however the rounding error in this case seems quite large to me; what I mean is I might have expected a result of 0.99990000001 or some similar 'close' result. And for the record I got the same answer in a JavaVM and in a .Net environment.

    Read the article

  • CPU Architecture and floating-point math

    - by Jo-Herman Haugholt
    I'm trying to wrap my head around some details about how floating point math is performed on the CPU, trying to better understand what data types to use etc. I think I have a fairly good understanding of how integer math is performed. If I've understood correctly, and disregarding SIMD, a 32-bit CPU will generally perform integer math at at least 32-bit precision etc. Is it correct that floating-point math is dependent on the presence of a FPU? And that the FPU on the x86 is 80-bit, so floating point math is performed at this precision unless using SIMD? What about ARM?

    Read the article

  • About floating point precision and why do we still use it

    - by system_is_b0rken
    Floating point has always been troublesome for precision on large worlds. This article explains behind-the-scenes and offers the obvious alternative - fixed point numbers. Some facts are really impressive, like: "Well 64 bits of precision gets you to the furthest distance of Pluto from the Sun (7.4 billion km) with sub-micrometer precision. " Well sub-micrometer precision is more than any fps needs (for positions and even velocities), and it would enable you to build really big worlds. My question is, why do we still use floating point if fixed point has such advantages? Most rendering APIs and physics libraries use floating point (and suffer it's disadvantages, so developers need to get around them). Are they so much slower? Additionally, how do you think scalable planetary engines like outerra or infinity handle the large scale? Do they use fixed point for positions or do they have some space dividing algorithm?

    Read the article

  • understanding floating point variables

    - by Syom
    There is some problem, i can't understand anyway. look at this code please <script type="text/javascript"> function math(x) { var y; y = x*10; alert(y); } </script> <input type="button" onclick="math(0.011)"> What must be alerted after i click on button? i think 0.11, but no, it alerts 0.10999999999999999 explain please this behavior. thanks in advance

    Read the article

  • how IEEE-754 floating point numbers work

    - by hatorade
    Let's say I have this: float i = 1.5 in binary, this float is represented as: 0 01111111 10000000000000000000000 I broke up the binary to represent the 'signed', 'exponent' and 'fraction' chunks. What I don't understand is how this represents 1.5. The exponent is 0 once you subtract the bias (127 - 127), and the fraction part with the implicit leading one is 1.1. How does 1.1 scaled by nothing = 1.5???

    Read the article

  • Find max integer size that a floating point type can handle without loss of precision

    - by Checkers
    Double has range more than a 64-bit integer, but its precision is less dues to its representation (since double is 64-bit as well, it can't fit more actual values). So, when representing larger integers, you start to lose precision in the integer part. #include <boost/cstdint.hpp> #include <limits> template<typename T, typename TFloat> void maxint_to_double() { T i = std::numeric_limits<T>::max(); TFloat d = i; std::cout << std::fixed << i << std::endl << d << std::endl; } int main() { maxint_to_double<int, double>(); maxint_to_double<boost::intmax_t, double>(); maxint_to_double<int, float>(); return 0; } This prints: 2147483647 2147483647.000000 9223372036854775807 9223372036854775800.000000 2147483647 2147483648.000000 Note how max int can fit into a double without loss of precision and boost::intmax_t (64-bit in this case) cannot. float can't even hold an int. Now, the question: is there a way in C++ to check if the entire range of a given integer type can fit into a loating point type without loss of precision? Preferably, it would be a compile-time check that can be used in a static assertion, and would not involve enumerating the constants the compiler should know or can compute.

    Read the article

  • floating point precision in ruby on rails model validations

    - by Chris Allison
    Hello I am trying to validate a dollar amount using a regex: ^[0-9]+\.[0-9]{2}$ This works fine, but whenever a user submits the form and the dollar amount ends in 0(zero), ruby(or rails?) chops the 0 off. So 500.00 turns into 500.0 thus failing the regex validation. Is there any way to make ruby/rails keep the format entered by the user, regardless of trailing zeros?

    Read the article

  • IE7 div floating bug

    - by Michael Frey
    I have the following <div id="border" style="width:100%; border:8px solid #FFF"> <div id="menu" style="width:250px; float:left;" > Some menu </div> <div id="content" style="padding-left:270px; width:520px;" > Main page content </div> </div> This gives me a left aligned menu and the content to the right of it, all surrounded by a border. On all browsers including IE8 it displays correctly. But on IE7 the content only starts below the menu, and leaves a big open space to the right of the menu. I have searched all kind of solutions and tried all kinds of combinations of right, left, none for float. clearing left right both. It always displays different on the browsers. Any help is appreciated. Michael

    Read the article

  • Menu floating to the right on IE and to the left in FF

    - by the_drow
    I am working on a website that has a menu which behaves correctly on FF but not on IE (as usuall). On IE it floats to the right while it should float to the left, however if float is set to none it behaves almost correctly, attaching the onto the top of the container. Here's the css: #navigation_wrap { background: url(../images/ltr/nav_bg.png); height: 34px; width: 954px; } .btn_login { float: right; margin: 4px 4px 0 0; } .navigation { float: left; } .navigation ul { list-style: none; margin: 8px 0 0 15px; } .navigation ul li { border-right: 1px solid white; float: left; padding: 0 12px 0 12px; } .navigation ul li.last { border: none; } .navigation ul li a { color: white; font-size: 14px; text-decoration: none; } .navigation ul li a:hover { text-decoration: underline; } .navigation ul li a.active { font-weight: bold; } And here's the html: <div id="navigation_wrap"> <div class="navigation"> <ul> <li><a class="active" href="default.asp">Home Page</a></li> <li><a class="" href="faq.asp">FAQ</a></li><li><a class="" href="articles.asp">Articles</a></li> <li><a class="" href="products.asp">Packages &amp; Pricing</a></li> <li><a class="" href="gp.asp?gpid=15">test1</a></li> <li><a class=" last" href="gp.asp?gpid=17">test asher</a></li> </ul> </div> <div class="btn_login"> ... </div> </div> I hope anyone would have an idea. Thanks, Omer.

    Read the article

  • How to make gcc on SUN calculate floating points the same way as in Linux

    - by Marina
    I have a project where I have to perform some mathematics calculations with double variables. The problem is that I get different results on SUN Solaris 9 and Linux. There are a lot of ways (explained here and other forums) how to make Linux work as Sun, but not the other way around. I cannot touch the Linux code, so it is only SUN I can change. Is there any way to make SUN to behave as Linux? The code I run(compile with gcc on both systems): int hash_func(char *long_id) { double product, lnum, gold; while (*long_id) lnum = lnum * 10.0 + (*long_id++ - '0'); printf("lnum => %20.20f\n", lnum); lnum = lnum * 10.0E-8; printf("lnum => %20.20f\n", lnum); gold = 0.6125423371582974; product = lnum * gold; printf("product => %20.20f\n", product); ... } if the input is 339886769243483 the output in Linux: lnum => 339886769243**483**.00000000000000000000 lnum => 33988676.9243**4829473495483398** product => 20819503.600158**59827399253845** When on SUN: lnum => 339886769243483.00000000000000000000 lnum => 33988676.92434830218553543091 product = 20819503.600158**60199928283691** Note: The result is not always different, moreover most of the times it is the same. Just 10 15-digit numbers out of 60000 have this problem. Please help!!!

    Read the article

  • how floating point numbers work in C

    - by hatorade
    Let's say I have this: float i = 1.5 in binary, this float is represented as: 0 01111111 10000000000000000000000 I broke up the binary to represent the 'signed', 'exponent' and 'fraction' chunks. What I don't understand is how this represents 1.5. The exponent is 0 once you subtract the bias (127 - 127), and the fraction part with the implicit leading one is 1.1. How does 1.1 scaled by nothing = 1.5???

    Read the article

  • Floating Point Arithmetic - Modulo Operator on Double Type

    - by CrimsonX
    So I'm trying to figure out why the modulo operator is returning such a large unusual value. If I have the code: double result = 1.0d % 0.1d; it will give a result of 0.09999999999999995. I would expect a value of 0 Note this problem doesn't exist using the dividing operator - double result = 1.0d / 0.1d; will give a result of 10.0, meaning that the remainder should be 0. Let me be clear: I'm not surprised that an error exists, I'm surprised that the error is so darn large compared to the numbers at play. 0.0999 ~= 0.1 and 0.1 is on the same order of magnitude as 0.1d and only one order of magnitude away from 1.0d. Its not like you can compare it to a double.epsilon, or say "its equal if its < 0.00001 difference". I've read up on this topic on StackOverflow, in the following posts one two three, amongst others. Can anyone suggest explain why this error is so large? Any any suggestions to avoid running into the problems in the future (I know I could use decimal instead but I'm concerned about the performance of that).

    Read the article

  • Rounding floating-point numbers to 4 decimal points

    - by Himadri
    I have two decimal numbers. I want those number to be same upto 4 decimal points without rounding. If numbers are different I want 2nd number to be replaced by 1st. What if condition should I write? Eg, 1. num1 = 0.94618976 num2 = 0.94620239 If we round these numbers upto 4 decimal then we get 0.9462 same number, but I don't want to round these numbers. 2. num1 = 0.94620239 num2 = 0.94639125 The one way I found is take absolute difference of both numbers say diff and then check the value. My problem is of checking the range of diff. Thank You.

    Read the article

  • Floating point precision and physics calculations

    - by Vee
    The gravity Vector2 in my physics world is (0; 0.1). The number 0.1 is known to be problematic, since "it cannot be represented exactly, but is approximately 1.10011001100110011001101 × 2-4". Having this value for the gravity gives me problems with collisions and creates quite nasty bugs. Changing the value to 0.11 solves these problems. Is there a more elegant solution that doesn't require changing the value at all?

    Read the article

  • Rejigging a floating point equation ...

    - by Jamie
    I'd like to know if there is a way to improve the accuracy of calculating a slope. (This came up a few months back here). It seems by changing: float get_slope(float dXa, float dXb, float dYa, float dYb) { return (dXa - dXb)/(dYa - dYb); } to float get_slope(float dXa, float dXb, float dYa, float dYb) { return dXa/(dYa - dYb) - dXb/(dYa - dYb); } might be an improvement. Suggestions? Edit: It's precision I'm after, not efficiency.

    Read the article

  • Floating point comparison in STL, BOOST

    - by Paul
    Is there in the STL or in Boost a set of generic simple comparison functions? The one I found are always requiring template parameters, and/or instantiation of a struct template. I'm looking for something with a syntax like : if ( is_greater(x,y) ) { ... } Which could be implemented as : template <typename T> bool is_greater(const T& x, const T& y) { return x > y + Precision<T>::eps; }

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >