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  • C# Web Reference and PHP

    - by Louis
    I am attempting to call a Web Service (created in PHP) from my C# application. I have successfully added the Web Reference in Visual Studios, however I cannot figure out exactly how to invoke the call to the web service. I have been following this tutorial: http://sanity-free.org/article25.html however when I try and compile I get a "The type or namespace name 'SimpleService' could not be found". My C# code is as follows: using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Text; using System.Windows.Forms; using System.IO; namespace inVision_OCR { public partial class Form1 : Form { public Form1() { InitializeComponent(); } private void translateButton_Click(object sender, EventArgs e) { // We need to snap the image here somehow . . . // Open up the image and read its contents into a string FileStream file = new FileStream("\\Hard Disk\\ocr_images\\test.jpg", FileMode.Open); StreamReader sr = new StreamReader(file); string s = sr.ReadToEnd(); sr.Close(); // Using SOAP, pass this message to our development server SimpleService svc = new SimpleService(); string s1 = svc.getOCR("test"); MessageBox.Show(s); } } } Any help would be appreciated.

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  • application-context.xml problem for Spring ActionScript

    - by jiri
    content of The application-content.xml is <?xml version="1.0" encoding="utf-8"?> <objects xmlns="http://www.springactionscript.org/schema/objects" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springactionscript.org/schema/objects http://www.springactionscript.org/schema/objects/spring-actionscript-objects-1.0.xsd"> <property file="strings.properties" /> <object id="string1" class="String"/> <object id="string2" class="String"/> <object id="nlslzf" class="com.sgb.model.MyClass"/> </objects> content of strings.properties is: s1=Firststring s2=Secondstring but error: Main Thread (Suspended: Error: An object definition for 'string1' was not found.) org.springextensions.actionscript.ioc.factory.support::AbstractObjectFactory/getObject SpringActionScript/test SpringActionScript/onCreationComplete SpringActionScript/___SpringActionScript_Application1_creationComplete flash.events::EventDispatcher/dispatchEventFunction [no source] mx.core::UIComponent/dispatchEvent mx.core::UIComponent/set initialized mx.managers::LayoutManager/doPhasedInstantiation Function/http://adobe.com/AS3/2006/builtin::apply [no source] mx.core::UIComponent/callLaterDispatcher2 mx.core::UIComponent/callLaterDispatcher2 mx.core::UIComponent/callLaterDispatcher i can run normal if removed the '' why it is? springactionscript bug?

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  • Signals and Variables in VHDL - Problem

    - by Morano88
    I have a signal and this signal is a bitvector. The length of the bitvector depends on an input n, it is not fixed. In order to find the length, I have to do some computations. Can I define a signal after defining the variables ? It is ggiving me errors when I do that. It is working fine If I keep the signal before the variables .. but I don't want that .. the length of Z depends on the computations of the variables. What is the solution ? library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL; use IEEE.STD_LOGIC_UNSIGNED.ALL; entity BSD_Full_Comp is Generic (n:integer:=8); Port(X, Y : inout std_logic_vector(n-1 downto 0); FZ : out std_logic_vector(1 downto 0)); end BSD_Full_Comp; architecture struct of BSD_Full_Comp is Component BSD_BitComparator Port ( Ai_1 : inout STD_LOGIC; Ai_0 : inout STD_LOGIC; Bi_1 : inout STD_LOGIC; Bi_0 : inout STD_LOGIC; S1 : out STD_LOGIC; S0 : out STD_LOGIC ); END Component; Signal Z : std_logic_vector(2*n-3 downto 0); begin ass : process Variable length : integer := n; Variable pow : integer :=0 ; Variable ZS : integer :=0; begin while length /= 0 loop length := length/2; pow := pow+1; end loop; length := 2 ** pow; ZS := length - n; wait; end process; end struct;

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  • Finding intersection of two spheres

    - by Onkar Deshpande
    Hi, Consider the following problem - I am given 2 links of length L0 and L1. P0 is the point that the first link starts at and P1 is the point that I want the end of second link to be at in 3-D space. I am supposed to write a function that should take in these 3-D points (P0 and P1) as inputs and should find all configurations of the links that put the second link's end point at P1. My understanding of how to go about it is - Each link L0 and L1 will create a sphere S0 and S1 around itself. I should find out the intersection of those two spheres (which will be a circle) and print all points that are on the circumference of that circle. I saw gmatt's first reply on the http://stackoverflow.com/questions/1406375/finding-intersection-points-between-3-spheres but could not understand it properly since the images did not show up. I also saw a formula for finding out the intersection at mathworld[dot]wolfram[dot]com/Sphere-SphereIntersection[dot]html . I could find the radius of intersection by the method given on mathworld. Also I can find the center of that circle and then use the parametric equation of circle to find the points. The only doubt that I have is will this method work for the points P0 and P1 mentioned above ? Please comment and let me know your thoughts.

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  • Best pattern to load enumerated values from DAL using WCF RIA Services

    - by Dale Halliwell
    I would like to be able to load several RIA entitysets in a single call without chaining/nesting several small LoadOperations together so that they load sequentially. I have several pages that have a number of comboboxes on them. These comboboxes are populated with static values from a database (for example status values). Right now I preload these values in my VM by one method that strings together a series of LoadOperations for each type that I want to load. For example: public void LoadEnums() { context.Load(context.GetMyStatusValues1Query()).Completed += (s, e) => { this.StatusValues1 = context.StatusValues1; context.Load(context.GetMyStatusValues2()).Completed += (s1, e1) => { this.StatusValues2 = context.StatusValues2; context.Load(context.GetMyStatusValues3Query()).Completed += (s2, e2) => { this.StatusValues3 = context.StatusValues3; (....and so on) }; }; }; }; While this works fine, it seems a bit nasty. Also, I would like to know when the last loadoperation completes so that I can load whatever entity I want to work on after this, so that these enumerated values resolve properly in form elements like comboboxes and listboxes. (I think) I can't do this easily above without creating a delegate and calling that on the completion of the last loadoperation. So my question is: does anyone out there know a better pattern to use, ideally where I can load all my static entitysets in a single LoadOperation?

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  • How do I iterate through the id properties of R.java class ?

    - by mgpyone
    I've 16 textViews and need to set something like this done: for (int i=1; i<6; i++) { int $RidInt = R.id.s; tv[i] = (TextView)findViewById($RidInt); tv[i].setTypeface(face); tv[i].setClickable(true); tv[i].setOnClickListener(clickListener); } my R.java file is : public final class R { public static final class attr { } public static final class drawable { public static final int icon=0x7f020000; } public static final class id { public static final int s1=0x7f050000; public static final int s10=0x7f050009; public static final int s11=0x7f05000a; public static final int s12=0x7f05000b; public static final int s13=0x7f05000c; public static final int s14=0x7f05000d; public static final int s15=0x7f05000e; public static final int s16=0x7f05000f; public static final int s2=0x7f050001; public static final int s3=0x7f050002; public static final int s4=0x7f050003; public static final int s5=0x7f050004; public static final int s6=0x7f050005; public static final int s7=0x7f050006; public static final int s8=0x7f050007; public static final int s9=0x7f050008; } public static final class layout { public static final int main=0x7f030000; public static final int toast=0x7f030001; } public static final class string { public static final int app_name=0x7f040000; public static final int s2=0x7f040001; } }

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  • Signals and Variables in VHDL (order) - Problem

    - by Morano88
    I have a signal and this signal is a bitvector (Z). The length of the bitvector depends on an input n, it is not fixed. In order to find the length, I have to do some computations. Can I define a signal after defining the variables ? It is giving me errors when I do that. It is working fine If I keep the signal before the variables (that what is showing below) .. but I don't want that .. the length of Z depends on the computations of the variables. What is the solution ? library IEEE; use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL; use IEEE.STD_LOGIC_UNSIGNED.ALL; entity BSD_Full_Comp is Generic (n:integer:=8); Port(X, Y : inout std_logic_vector(n-1 downto 0); FZ : out std_logic_vector(1 downto 0)); end BSD_Full_Comp; architecture struct of BSD_Full_Comp is Component BSD_BitComparator Port ( Ai_1 : inout STD_LOGIC; Ai_0 : inout STD_LOGIC; Bi_1 : inout STD_LOGIC; Bi_0 : inout STD_LOGIC; S1 : out STD_LOGIC; S0 : out STD_LOGIC ); END Component; Signal Z : std_logic_vector(2*n-3 downto 0); begin ass : process Variable length : integer := n; Variable pow : integer :=0 ; Variable ZS : integer :=0; begin while length /= 0 loop length := length/2; pow := pow+1; end loop; length := 2 ** pow; ZS := length - n; wait; end process; end struct;

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  • How do I iterate through the id properties of R.java in android?

    - by mgpyone
    I've 16 textViews and need to set something like this done: for (int i=1; i<6; i++) { int $RidInt = R.id.s; tv[i] = (TextView)findViewById($RidInt); tv[i].setTypeface(face); tv[i].setClickable(true); tv[i].setOnClickListener(clickListener); } my R.java file is : public final class R { public static final class attr { } public static final class drawable { public static final int icon=0x7f020000; } public static final class id { public static final int s1=0x7f050000; public static final int s10=0x7f050009; public static final int s11=0x7f05000a; public static final int s12=0x7f05000b; public static final int s13=0x7f05000c; public static final int s14=0x7f05000d; public static final int s15=0x7f05000e; public static final int s16=0x7f05000f; public static final int s2=0x7f050001; public static final int s3=0x7f050002; public static final int s4=0x7f050003; public static final int s5=0x7f050004; public static final int s6=0x7f050005; public static final int s7=0x7f050006; public static final int s8=0x7f050007; public static final int s9=0x7f050008; } public static final class layout { public static final int main=0x7f030000; public static final int toast=0x7f030001; } public static final class string { public static final int app_name=0x7f040000; public static final int s2=0x7f040001; } }

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  • Using Hidden Markov Model for designing AI mp3 player

    - by Casper Slynge
    Hey guys. Im working on an assignment, where I want to design an AI for a mp3 player. The AI must be trained and designed with the use of a HMM method. The mp3 player shall have the functionality of adapting to its user, by analyzing incoming biological sensor data, and from this data the mp3 player will choose a genre for the next song. Given in the assignment is 14 samples of data: One sample consist of Heart Rate, Respiration, Skin Conductivity, Activity and finally the output genre. Below is the 14 samples of data, just for you to get an impression of what im talking about. Sample HR RSP SC Activity Genre S1 Medium Low High Low Rock S2 High Low Medium High Rock S3 High High Medium Low Classic S4 High Medium Low Medium Classic S5 Medium Medium Low Low Classic S6 Medium Low High High Rock S7 Medium High Medium Low Classic S8 High Medium High Low Rock S9 High High Low Low Classic S10 Medium Medium Medium Low Classic S11 Medium Medium High High Rock S12 Low Medium Medium High Classic S13 Medium High Low Low Classic S14 High Low Medium High Rock My time of work regarding HMM is quite low, so my question to you is if I got the right angle on the assignment. I have three different states for each sensor: Low, Medium, High. Two observations/output symbols: Rock, Classic In my own opinion I see my start probabilities as the weightened factors for either a Low, Medium or High state in the Heart Rate. So the ideal solution for the AI is that it will learn these 14 sets of samples. And when a users sensor input is received, the AI will compare the combination of states for all four sensors, with the already memorized samples. If there exist a matching combination, the AI will choose the genre, and if not it will choose a genre according to the weightened transition probabilities, while simultaniously updating the transition probabilities with the new data. Is this a right approach to take, or am I missing something ? Is there another way to determine the output probability (read about Maximum likelihood estimation by EM, but dont understand the concept)? Best regards, Casper

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  • How to use the same element name in different purposes ( in XML and DTD ) ?

    - by BugKiller
    Hi, I Want to create a DTD schema for this xml document: <root> <student> <name> <firstname>S1</firstname> <lastname>S2</lastname> </name> </student> <course> <name>CS101</name> </course> </root> as you can see , the element name in the course contains plain text ,but the element name in the student is complex type ( first-name, last-name ). The following is the DTD: <!ELEMENT root (course|student)*> <!ELEMENT student (name)> <!ELEMENT name (lastname|firstname)> <!ELEMENT firstname (#PCDATA)> <!ELEMENT lastname (#PCDATA)> <!ELEMENT course (name)> When I want to validate it , I get an error because the course's name has different structure then the student's name . My Question: how can I make a work-around solution for this situation without changing the name of element name using DTD not xml schema . Thanks.

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  • c# calling process "cannot find the file specified"

    - by laura
    I'm a c# newbie so bear with me. I'm trying to call "pslist" from PsTools from a c# app, but I keep getting "The system cannot find the file specified". I thought I read somewhere on google that the exe should be in c:\windows\system32, so I tried that, still nothing. Even trying the full path to c:\windows\system32\PsList.exe is not working. I can open other things like notepad or regedit. Any ideas? C:\WINDOWS\system32dir C:\WINDOWS\SYSTEM32\PsList.exe Volume in drive C has no label. Volume Serial Number is ECC0-70AA Directory of C:\WINDOWS\SYSTEM32 04/27/2010 11:04 AM 231,288 PsList.exe 1 File(s) 231,288 bytes 0 Dir(s) 8,425,492,480 bytes free try { // Start the child process. Process p = new Process(); // Redirect the output stream of the child process. p.StartInfo.UseShellExecute = false; p.StartInfo.RedirectStandardOutput = true; //This works //p.StartInfo.FileName = @"C:\WINDOWS\regedit.EXE"; //This doesn't p.StartInfo.FileName = @"C:\WINDOWS\system32\PsList.exe"; p.Start(); // Do not wait for the child process to exit before // reading to the end of its redirected stream. p.WaitForExit(); // Read the output stream first and then wait. s1 = p.StandardOutput.ReadToEnd(); p.WaitForExit(); } catch (Exception ex) { Console.WriteLine("Exception Occurred :{0},{1}", ex.Message, ex.StackTrace.ToString()); Console.ReadLine(); }

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  • propositional theorems

    - by gcc
    wang's theorem to prove theorems in propositonal calculus Label Sequent Comment S1: P ? Q, Q ? R, ¬R ? ¬P Initial sequent. S2: ¬P ? Q, ¬Q ? R, ¬R ? ¬P Two applications of R5. S3: ¬P ? Q, ¬Q ? R ? ¬P, R Rl. S4: ¬P, ¬Q ? R ? ¬P, R S4 and S5 are obtained from S3 with R3. Note that S4 is an axiom since P appears on both sides of the sequent ar- row at the top level. S5: Q, ¬Q ? R ? ¬P, R The other sequent generated by the ap- plication of R3. S6: Q, ¬Q ? ¬P, R S6 and S7 are obtained from S5 using R3. S7: Q, R ? ¬P, R This is an axiom. S8: Q ? ¬P, R, Q Obtained from S6 using R1. S8 is an axiom. The original sequent is now proved, since it has successfully been transformed into a set of three axioms with no unproved sequents left over. //R1,R2,R3,R4,R5 is transformation myhomework is so complicate as you can see and i must finish homework in 4 days i just want help ,can you show me how i should construct the algorithm or how i should take and store the input

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  • How to use the same element name for different purposes ( in XML and DTD ) ?

    - by BugKiller
    Hi, I Want to create a DTD schema for this xml document: <root> <student> <name> <firstname>S1</firstname> <lastname>S2</lastname> </name> </student> <course> <name>CS101</name> </course> </root> as you can see , the element name in the course contains plain text ,but the element name in the student is complex type ( first-name, last-name ). The following is the DTD: <!ELEMENT root (course|student)*> <!ELEMENT student (name)> <!ELEMENT name (lastname|firstname)> <!ELEMENT firstname (#PCDATA)> <!ELEMENT lastname (#PCDATA)> <!ELEMENT course (name)> When I want to validate it , I get an error because the course's name has different structure then the student's name . My Question: how can I make a work-around solution for this situation without changing the name of element name using DTD not xml schema . Thanks.

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  • jQuery: show an element from select drop down, hide it when other option selected

    - by Ricardo Zea
    I've tried looking around and there are similar problems, but mine is way more simple, but yet, I can't find a solution within these forums. While learning jQuery, I'm trying to show a DIV when an item/option from a select drop down is selected, and hide that same DIV when any other option in the select drop down is selected. select HTML: <select name="source" id="source"> <option value="null" selected="selected">&mdash;Select&mdash;</option> <option value="s1">Source 1</option> <option value="s2">Source 2</option> <option value="sother">Other</option> </select> DIV I need to show when 'Other' is selected: <div id="specify-source">Other source here...</div> When any other option in the select menu is selected, the above DIV shouldn't be visible. I've tried this jQuery but of course it doesn't work properly: $(function() { $.viewMap = { 'sother' : $('#specify-source') }; $('#source').change(function() { // hide all $.each($.viewMap, function() { this.hide(); }); // show current $.viewMap[$(this).val()].show(); }); }); Any help you can give me, I'd greatly appreciate it. Thanks,

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  • Splitting a set of object into several subsets of 'similar' objects

    - by doublep
    Suppose I have a set of objects, S. There is an algorithm f that, given a set S builds certain data structure D on it: f(S) = D. If S is large and/or contains vastly different objects, D becomes large, to the point of being unusable (i.e. not fitting in allotted memory). To overcome this, I split S into several non-intersecting subsets: S = S1 + S2 + ... + Sn and build Di for each subset. Using n structures is less efficient than using one, but at least this way I can fit into memory constraints. Since size of f(S) grows faster than S itself, combined size of Di is much less than size of D. However, it is still desirable to reduce n, i.e. the number of subsets; or reduce the combined size of Di. For this, I need to split S in such a way that each Si contains "similar" objects, because then f will produce a smaller output structure if input objects are "similar enough" to each other. The problems is that while "similarity" of objects in S and size of f(S) do correlate, there is no way to compute the latter other than just evaluating f(S), and f is not quite fast. Algorithm I have currently is to iteratively add each next object from S into one of Si, so that this results in the least possible (at this stage) increase in combined Di size: for x in S: i = such i that size(f(Si + {x})) - size(f(Si)) is min Si = Si + {x} This gives practically useful results, but certainly pretty far from optimum (i.e. the minimal possible combined size). Also, this is slow. To speed up somewhat, I compute size(f(Si + {x})) - size(f(Si)) only for those i where x is "similar enough" to objects already in Si. Is there any standard approach to such kinds of problems? I know of branch and bounds algorithm family, but it cannot be applied here because it would be prohibitively slow. My guess is that it is simply not possible to compute optimal distribution of S into Si in reasonable time. But is there some common iteratively improving algorithm?

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  • byte and short data types in Java can accept the value outside the range by explicit cast. The higher data types however can not. Why?

    - by Lion
    Let's consider the following expressions in Java. byte a = 32; byte b = (byte) 250; int i = a + b; This is valid in Java even though the expression byte b = (byte) 250; is forced to assign the value 250 to b which is outside the range of the type byte. Therefore, b is assigned -6 and consequently i is assigned the value 26 through the statement int i = a + b;. The same thing is possible with short as follows. short s1=(short) 567889999; Although the specified value is outside the range of short, this statement is legal. The same thing is however wrong with higher data types such int, double, folat etc and hence, the following case is invalid and causes a compile-time error. int z=2147483648; This is illegal, since the range of int in Java is from -2,147,483,648 to 2147483647 which the above statement exceeds and issues a compile-time error. Why is such not wrong with byte and short data types in Java?

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  • Help with string equality in Java

    - by annayena
    The following function accepts 2 strings, the 2nd (not 1st) possibly containing *'s (asterisks). An * is a replacement for a string (empty, 1 char or more), it can appear appear (only in s2) once, twice, more or not at all, it cannot be adjacent to another * (ab**c), no need to check that. public static boolean samePattern(String s1, String s2) It returns true if strings are of the same pattern. It must be recursive, not use any loops, static or global variables. Also it's prohibited to use the method equals in the String class. Can use local variables and method overloading. Can use only these methods: charAt(i), substring(i), substring(i, j), length(). Examples: 1: TheExamIsEasy; 2: "The*xamIs*y" ---> true 1: TheExamIsEasy; 2: "Th*mIsEasy*" ---> true 1: TheExamIsEasy; 2: "*" ---> true 1: TheExamIsEasy; 2: "TheExamIsEasy" ---> true 1: TheExamIsEasy; 2: "The*IsHard" ---> FALSE I am stucked on this question for many hours now! I need the solution in Java please kindly help me.

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  • c# logic to get the first non-repeating(distinct) character from the string

    - by NoviceToDotNet
    In c# i want to create logic that if i a string like abcabda is passed to a method then it should return first non repeative character from string like in above it should return c. i am unable to convert a string to array of character then how to make comparison of each array character to the string and return the first non repeative character. CanI make it like this? class A { static void main() { A a=new A(); char ch=a.m1(abcabd); } } class B { char m1(string s) { string s1=s; char[] ch1=new char[s.length]; for(int x=0; x<s.length;x++) { ch1[x]=s[x]; } for(int x=0; x<s.length; x++) { for(int y=0; y<s.lenth; y++) { if(s[x]=ch1[y]) { /// here i am confused how to create logic for comparison please let me know // and how to return the character } } } } }

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  • HELP IN JAVA!! URGENT

    - by annayena
    The following function accepts 2 strings, the 2nd (not 1st) possibly containing *'s (asterisks). An * is a replacement for a string (empty, 1 char or more), it can appear appear (only in s2) once, twice, more or not at all, it cannot be adjacent to another * (ab**c), no need to check that. public static boolean samePattern(String s1, String s2) It returns true if strings are of the same pattern. It must be recursive, not use any loops, static & global variables. Also it's PROHIBITED to use the method equals in the String class. Can use local variables & method overloading. Can use only these methods: charAt(i), substring(i), substring(i, j), length(). Examples: 1: TheExamIsEasy; 2: "The*xamIs*y" --- true 1: TheExamIsEasy; 2: "Th*mIsEasy*" --- true 1: TheExamIsEasy; 2: "*" --- true 1: TheExamIsEasy; 2: "TheExamIsEasy" --- true 1: TheExamIsEasy; 2: "The*IsHard" --- FALSE I am stucked on this question for many hours now! I need the solution in Java please kindly help me.

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  • Unable to checkout svn repositories

    - by lucaghera
    I have an ubuntu 12.04 machine were apache2 is set up with SSL certificates. In the same machine there is a SVN server. It all worked great till the update to 12.04. Now I'm able to access the svn via a web-browser and also by using an eclipse plugin (subversive), but I'm not able to access the svn via command line. When I try to check out a repo from a Mac Os X client it returns: svn: E120171: Unable to connect to a repository at URL 'https://IP/svn/repo_name' svn: E120171: Error running context: An error occurred during SSL communication If I try to check out a repo from an Ubuntu client it returns: svn: OPTIONS of 'https://IP/svn/repo_name': SSL handshake failed: SSL error: A TLS warning alert has been received. (https://IP)

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  • Adobe Photoshop Vs Lightroom Vs Aperture

    - by Aditi
    Adobe Photoshop is the standard choice for photographers, graphic artists and Web designers. Adobe Photoshop Lightroom  & Apple’s Aperture are also in the same league but the usage is vastly different. Although Photoshop is most popular & widely used by photographers, but in many ways it’s less relevant to photographers than ever before. As Lightroom & Aperture is aimed squarely at photographers for photo-processing. With this write up we are going to help you choose what is right for you and why. Adobe Photoshop Adobe Photoshop is the most liked tool for the detailed photo editing & designing work. Photoshop provides great features for rollover and Image slicing. Adobe Photoshop includes comprehensive optimization features for producing the highest quality Web graphics with the smallest possible file sizes. You can also create startling animations with it. Designers & Editors know how important precise masking is, PhotoShop lets you do that with various detailing tools. Art history brush, contact sheets, and history palette are some of the smart features, which add to its viability. Download Whether you’re producing printed pages or moving images, you can work more efficiently and produce better results because of its smooth integration across other adobe applications. Buy supporting layer effects, it allows you to quickly add drop shadows, inner and outer glows, bevels, and embossing to layers. It also provides Seamless Web Graphics Workflow. Photoshop is hands-down the BEST for editing. Photoshop Cons: • Slower, less precise editing features in Bridge • Processing lots of images requires actions and can be slower than exporting images from Lightroom • Much slower with editing and processing a large number of images Aperture Apple Aperture is aimed at the professional photographer who shoots predominantly raw files. It helps them to manage their workflow and perform their initial Raw conversion in a better way. Aperture provides adjustment tools such as Histogram to modify color and white balance, but most of the editing of photos is left for Photoshop. It gives users the option of seeing their photographs laid out like slides or negatives on a light table. It boasts of – stars, color-coding and easy techniques for filtering and picking images. Aperture has moved forward few steps than Photoshop, but most of the editing work has been left for Photoshop as it features seamless Photoshop integration. Aperture Pros: Aperture is a step up from the iPhoto software that comes with every Mac, and fairly easy to learn. Adjustments are made in a logical order from top to bottom of the menu. You can store the images in a library or any folder you choose. Aperture also works really well with direct Canon files. It is just $79 if you buy it through Apple’s App Store Moving forward, it will run on the iPad, and possibly the iPhone – Adobe products like Lightroom and Photoshop may never offer these options It is much nicer and simpler user interface. Lightroom Lightroom does a smashing job of basic fixing and editing. It is more advanced tool for photographers. They can use it to have a startling photography effect. Light room has many advanced features, which makes it one of the best tools for photographers and far ahead of the other two. They are Nondestructive editing. Nothing is actually changed in an image until the photo is exported. Better controls over organizing your photos. Lightroom helps to gather a group of photos to use in a slideshow. Lightroom has larger Compare and Survey views of images. Quickly customizable interface. Simple keystrokes allow you to perform different All Lightroom controls are kept available in panels right next to the photos. Always-available History palette, it doesn’t go when you close lightroom. You gain more colors to work with compared to Photoshop and with more precise control. Local control, or adjusting small parts of a photo without affecting anything else, has long been an important part of photography. In Lightroom 2, you can darken, lighten, and affect color and change sharpness and other aspects of specific areas in the photo simply by brushing your cursor across the areas. Photoshop has far more power in its Cloning and Healing Brush tools than Lightroom, but Lightroom offers simple cloning and healing that’s nondestructive. Lightroom supports the RAW formats of more cameras than Aperture. Lightroom provides the option of storing images outside the application in the file system. It costs less than photoshop. Download Why PhotoShop is advanced than Lightroom? There are countless image processing plug-ins on the market for doing specialized processing in Photoshop. For example, if your image needs sophisticated noise reduction, you can use the Noiseware plug-in with Photoshop to do a much better job or noise removal than Lightroom can do. Lightroom’s advantages over Aperture 3 Will always have better integration with Photoshop. Lightroom is backed by bigger and more active user community (So abundant availability for tutorials, etc.) Better noise reduction tool. Especially for photographers the Lens-distortion correction tool  is perfect Lightroom Cons: • Have to Import images to work on them • Slows down with over 10,000 images in the catalog • For processing just one or two images this is a slower workflow Photoshop Pros: • ACR has the same RAW processing controls as Lightroom • ACR Histogram is specialized to the chosen color space (Lightroom is locked into ProPhoto RGB color space with an sRGB tone curve) • Don’t have to Import images to open in Bridge or ACR • Ability to customize processing of RAW images with Photoshop Actions Pricing and Availability Get LightRoomGet PhotoShop Latest version Of Photoshop can be purchased from Adobe store and Adobe authorized reseller and it costs US$999. Latest version of Aperture can be bought for US$199 from Apple Online store or Mac App Store. You can buy latest version of LightRoom from Adobe Store or Adobe Authorized reseller for US$299. 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  • The Beginner’s Guide to Managing Users and Groups in Linux

    - by Zainul Franciscus
    Ubuntu Linux uses groups to help you manage users, set permissions on those users, and even monitor how much time they are spending in front of the PC. Here’s a beginner’s guide to how it all works Latest Features How-To Geek ETC The How-To Geek Holiday Gift Guide (Geeky Stuff We Like) LCD? LED? Plasma? The How-To Geek Guide to HDTV Technology The How-To Geek Guide to Learning Photoshop, Part 8: Filters Improve Digital Photography by Calibrating Your Monitor Our Favorite Tech: What We’re Thankful For at How-To Geek The How-To Geek Guide to Learning Photoshop, Part 7: Design and Typography Happy Snow Bears Theme for Chrome and Iron [Holiday] Download Full Command and Conquer: Tiberian Sun Game for Free Scorched Cometary Planet Wallpaper Quick Fix: Add the RSS Button Back to the Firefox Awesome Bar Dropbox Desktop Client 1.0.0 RC for Windows, Linux, and Mac Released Hang in There Scrat! – Ice Age Wallpaper

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  • What PC for programming? [on hold]

    - by James Jeffery
    I'm asking this here because I'm looking for some advice on a PC that will be suitable for my needs. I currently have mac's and have rarely used PC's apart from my Vaio laptop, which is on it's way out. I will be using the PC for C# and .NET development. I mainly develop desktop apps using a PC, but I will be doing some ASP.NET as I'm switching from PHP to ASP. The selection of PC's are on here: http://www.pcworld.co.uk/ I have £500, but if I can not spend all of that I'd be happy. I will be doing nothing on the computer apart from C# development (desktop and ASP). Any help would be much appreciated. My applications are not intensive. They are usually automation software for web scraping and marketing purposes.

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  • How to Access a Windows Desktop From Your Tablet or Phone

    - by Chris Hoffman
    iPads and Android tablets can’t run Windows apps locally, but they can access a Windows desktops remotely — even with a physical keyboard. In a pinch, the same tricks can be used to access a Windows desktop from a smartphone. Microsoft recently launched their own official Remote Desktop app for iOS and Android devices. Microsoft’s official apps are primarily useful for businesses — if you’re a typical home user, you’ll want to use a different remote desktop solution. Microsoft’s Remote Desktop App Microsoft now offers official Remote Desktop apps for iPad and iPhone as well as Android tablets and smartphones. The apps use Microsoft’s RDP protocol to connect to remote Windows systems. They’re essentially just new clients for the Remote Desktop feature that has been included in Windows for more than a decade. There are big problems with these apps if you’re an average home user. Microsoft’s Remote Desktop server is not available on standard or Home versions of Windows, only Professional and Enterprise editions. If you do have the appropriate edition of Windows, you’ll have to set up port-forwarding and a dynamic DNS service if you want to access your Windows desktop from outside your local network. You could also set up a VPN — either way you’ll need to do some footwork. This app is a gift to businesses who are already using Remote Desktop and enthusiasts who have the more expensive versions of Windows and don’t mind the configuration process. To set this up, follow our guide to setting up Remote Desktop for Internet access and connect using the Remote Desktop app instead of traditional Remote Desktop clients. TeamViewer If you have the standard edition of Windows or you just don’t want to mess around with port-forwarding and dynamic DNS configuration, you’ll want to skip Remote Desktop and use something else. We like TeamViewer for this. Just as it’s a great way to remotely troubleshoot your relatives’ computers, it’s also a great way to remotely access your own computer. It doesn’t have the same limitations Microsoft’s Remote Desktop system has — it’s completely free for personal use, runs on any edition of Windows, and is easy to set up. There’s no messing around with port-forwarding or dynamic DNS configuration. To get started, just download and run the TeamViewer program on your computer. You can get started with it immediately, but you’ll want to set up unattended access to connect remotely without using the codes displayed on your screen. To connect, just install the TeamViewer mobile app and log in with the details the TeamViewer window displays. TeamViewer also offers software that runs on Mac and Linux, so you can remote-control other types of computers from your tablet. Other Options Microsoft’s Remote Desktop app and TeamViewer aren’t the only options, of course. There are a variety of different apps and services built for this. Splashtop is another fairly popular remote desktop solution that some people report as being faster. Unfortunately, it’s not entirely free — the iPad and iPhone app costs $20 at regular price. To use it over the Internet, you’ll have to purchase an additional “Anywhere Access Pack.” If you’re frustrated with TeamViewer’s speed and you don’t mind spending money, you may want to try Splashtop instead. As always, you could use any VNC server along with a VNC client app. VNC is the do-it-yourself solution — it’s an open protocol. Unlike Microsoft’s RDP protocol, you can install a VNC server of your own, configure it how you like, and use any mobile VNC client app. This is more flexible because you can install a VNC server on any edition of Windows or even non-Windows operating systems, but it otherwise has all the same issues — you have to worry about port-forwarding, setting up dynamic DNS, and securing your VNC server. Keep an eye on Chrome Remote Desktop. Chrome already offers a built-in remote desktop feature that allows you to remotely control your PC from another Windows, Mac, Linux, or Chrome OS device. Google is rumored to be building an Android app for Chrome Remote Desktop, which would allow you to easily access a computer running Chrome from Android tablets. Google’s solution is much more user-friendly for average people than Microsoft’s Remote Desktop solution, which is clearly geared towards businesses. Chrome Remote Desktop just requires signing in with a Google account. Remote desktop solutions like Microsoft’s Remote Desktop app and TeamViewer are also available for Windows tablets. On Windows RT devices like the Surface RT and Surface 2, they allow you to use the full Windows desktop that’s unavailable on your tablet.     

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  • This Week in Geek History: NORAD Tracks Santa, First HTTP Test, Babbage’s Birthday

    - by Jason Fitzpatrick
    History trivia shouldn’t be limited to just treaty dates and wars ending, we’re marking off major milestones in geek history—one week at at time. This week in history we’ve got Santa on the Cold War radar, baby HTTP going for a spin, and Babbage’s birth to help usher in the age of computers. Latest Features How-To Geek ETC How to Use the Avira Rescue CD to Clean Your Infected PC The Complete List of iPad Tips, Tricks, and Tutorials Is Your Desktop Printer More Expensive Than Printing Services? 20 OS X Keyboard Shortcuts You Might Not Know HTG Explains: Which Linux File System Should You Choose? HTG Explains: Why Does Photo Paper Improve Print Quality? An Alternate Star Wars Christmas Special [Video] Sunset in a Tropical Paradise Wallpaper Natural Wood Grain Icons for Your Desktop and App Launcher Docks My Blackberry Is Not Working! The Apple Too?! [Funny Video] Hidden Tracks Your Stolen Mac; Free Until End of January Why the Other Checkout Line Always Moves Faster

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