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  • ShowPlan Operator of the Week - Merge Join

    Did you ever wonder how and why your indexes affect the performances of joins? Once you've read Fabiano Amorim's unforgettable explanation, you'll learn to love the MERGE operator, and plan your indexes so as to allow the Query Optimiser to use it. Free trial of SQL Backup™“SQL Backup was able to cut down my backup time significantly AND achieved a 90% compression at the same time!” Joe Cheng. Download a free trial now.

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  • Cannot use ternary operator in LINQ query

    - by Nissan Fan
    I can't figure out why I get a Object reference not set to an instance of an object. error if I use a ternary operator in my LINQ query. var courses = from d in somesource orderby d.SourceName, d.SourceType select new { ID = d.InternalCode, Name = string.Format("{0} - {1}{2}", d.InternalCode, d.SourceName, (d.SourceType.Length > 0 ? ", " + d.SourceType : string.Empty)) }; Any thoughts?

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  • Call the cast operator of template base class within the derived class

    - by yoni
    I have a template class, called Cell, here the definition: template <class T> class OneCell { ..... } I have a cast operator from Cell to T, here virtual operator const T() const { ..... } Now i have derived class, called DCell, here template <class T> class DCell : public Cell<T> { ..... } I need to override the Cell's cast operator (insert a little if), but after I need to call the Cell's cast operator. In other methods it's should be something like virtual operator const T() const { if (...) { return Cell<T>::operator const T; } else throw ... } but i got a compiler error error: argument of type 'const int (Cell::)()const' does not match 'const int' What can I do? Thank you, and sorry about my poor English.

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  • What are the best practices for implementing the == operator for a class in C#?

    - by remio
    While implementing an == operator, I have the feeling that I am missing some essential points. Hence, I am searching some best practices around that. Here are some related questions I am thinking about: How to cleanly handle the reference comparison? Should it be implemented through a IEquatable<T>-like interface? Or overriding object.Equals? And what about the != operator? (this list might not be exhaustive).

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  • How Does iPhone Visual Voicemail Work From An Operator Perspective?

    - by Jasarien
    I'm hoping there are some Cell Phone Operator gurus here today. Would anyone be able to explain how Operators achieve the Visual Voicemail feature on the iPhone (and I assume other newer smart phones)? If a new cell phone operator that distributed SIM cards wanted to utilise the visual voicemail feature on unlocked iPhone's what services need to be in place to be able to support it? Is there an open spec or is it completely proprietary?

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  • "AND Operator" in PAM

    - by d_inevitable
    I need to prevent users from authenticating through Kerberos when the encrypted /home/users has not yet been mounted. (This is to avoid corrupting the ecryptfs mountpoint) Currently I have these lines in /etc/pam.d/common-auth: auth required pam_group.so use_first_pass auth [success=2 default=ignore] pam_krb5.so minimum_uid=1000 try_first_pass auth [success=1 default=ignore] pam_unix.so nullok_secure try_first_pass I am planning to use pam_exec.so to execute a script that will exit 1 if the ecyptfs mounts are not ready yet. Doing this: auth required pam_exec.so /etc/security/check_ecryptfs will lock me out for good if ecryptfs for some reason fails. In such case I would like to at least be able to login with a local (non-kerberos) user to fix the issue. Is there some sort of AND-Operator in which I can say that login through kerberos+ldap is only sufficient if both kerberos authentication and the ecryptfs mount has succeeded?

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  • Avoid Postfix Increment Operator

    - by muntoo
    I've read that I should avoid the postfix increment operator because of performance reasons (in certain cases). But doesn't this affect code readability? In my opinion: for(int i = 0; i < 42; i++); /* i will never equal 42! */ Looks better than: for(int i = 0; i < 42; ++i); /* i will never equal 42! */ But this is probably just out of habit. Admittedly, I haven't seen many use ++i. Is the performance that bad to sacrifice readability, in this case? Or am I just blind, and ++i is more readable than i++?

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  • Why does virtual assignment behave differently than other virtual functions of the same signature?

    - by David Rodríguez - dribeas
    While playing with implementing a virtual assignment operator I have ended with a funny behavior. It is not a compiler glitch, since g++ 4.1, 4.3 and VS 2005 share the same behavior. Basically, the virtual operator= behaves differently than any other virtual function with respect to the code that is actually being executed. struct Base { virtual Base& f( Base const & ) { std::cout << "Base::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Base::operator=(Base const &)" << std::endl; return *this; } }; struct Derived : public Base { virtual Base& f( Base const & ) { std::cout << "Derived::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Derived::operator=( Base const & )" << std::endl; return *this; } }; int main() { Derived a, b; a.f( b ); // [0] outputs: Derived::f(Base const &) (expected result) a = b; // [1] outputs: Base::operator=(Base const &) Base & ba = a; Base & bb = b; ba = bb; // [2] outputs: Derived::operator=(Base const &) Derived & da = a; Derived & db = b; da = db; // [3] outputs: Base::operator=(Base const &) ba = da; // [4] outputs: Derived::operator=(Base const &) da = ba; // [5] outputs: Derived::operator=(Base const &) } The effect is that the virtual operator= has a different behavior than any other virtual function with the same signature ([0] compared to [1]), by calling the Base version of the operator when called through real Derived objects ([1]) or Derived references ([3]) while it does perform as a regular virtual function when called through Base references ([2]), or when either the lvalue or rvalue are Base references and the other a Derived reference ([4],[5]). Is there any sensible explanation to this odd behavior?

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  • operator new for array of class without default constructor......

    - by skydoor
    For a class without default constructor, operator new and placement new can be used to declare an array of such class. When I read the code in More Effective C++, I found the code as below(I modified some part)..... My question is, why [] after the operator new is needed? I test it without it, it still works. Can any body explain that? class A { public: int i; A(int i):i(i) {} }; int main() { void *rawMemory = operator new[] (10 * sizeof(A)); // Why [] needed here? A *p = static_cast<A*>(rawMemory); for(int i = 0 ; i < 10 ; i++ ) { new(&p[i])A(i); } for(int i = 0 ; i < 10 ; i++ ) { cout<<p[i].i<<endl; } for(int i = 0 ; i < 10 ; i++ ) { p[i].~A(); } return 0; }

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  • Where to add an overloaded operator for the tr1::array?

    - by phlipsy
    Since I need to add an operator& for the std::tr1::array<bool, N> I wrote the following lines template<std::size_t N> std::tr1::array<bool, N> operator& (const std::tr1::array<bool, N>& a, const std::tr1::array<bool, N>& b) { std::tr1::array<bool, N> result; std::transform(a.begin(), a.end(), b.begin(), result.begin(), std::logical_and<bool>()); return result; } Now I don't know in which namespace I've to put this function. I considered the std namespace as a restricted area. Only total specialization and overloaded function templates are allowed to be added by the user. Putting it into the global namespace isn't "allowed" either in order to prevent pollution of the global namespace and clashes with other declarations. And finally putting this function into the namespace of the project doesn't work since the compiler won't find it there. What had I best do? I don't want to write a new array class putted into the project namespace. Because in this case the compiler would find the right namespace via argument dependent name lookup. Or is this the only possible way because writing a new operator for existing classes means extending their interfaces and this isn't allowed either for standard classes?

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  • How to replace a codebehind method with an aspx method using ternary operator

    - by user466663
    I have an asp:hyperlink control as part of a gridview template. The code in the aspx page is given below: asp:HyperLink runat="server" ID="lnkEdit" ToolTip="Edit article" NavigateUrl='<%# GetUrl(Eval("ID").ToString(), Eval("CategoryID").ToString()) %' ImageUrl="~/Images/Edit.gif" The NavigateUrl value is obtained from the codebehind method GetUrl(string, string). The code works fine and is as follows: protected string GetUrl(string id, string categoryID) { var CategoryID = string.Empty; if (!String.IsNullOrEmpty(Request.QueryString["CatID"])) { CategoryID = Request.QueryString["CatID"].ToString(); } else if (!String.IsNullOrEmpty(categoryID)) { CategoryID = categoryID; } return "~/TBSArticles/WriteOrEditArticle.aspx?ID=" + id + "&CatID=" + CategoryID; } I want to replace the code behind method by using a ternary operator within the aspx page. I tried something like below, but didn't work: asp:HyperLink runat="server" ID="lnkEdit" ToolTip="Edit article" NavigateUrl='<%# "~/TBSArticles/WriteOrEditArticle.aspx?ID=" + Eval("ID") + "&CatID=" + Eval(this.Request.QueryString["CatID"].ToString()) != ""? this.Request.QueryString["CatID"] : Eval("CategoryID")) %' ImageUrl="~/Images/Edit.gif" Any help will be greatly appreciated. Thanks

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  • Operator overloading in generic struct: can I create overloads for specific kinds(?) of generic?

    - by Carson Myers
    I'm defining physical units in C#, using generic structs, and it was going okay until I got the error: One of the parameters of a binary operator must be the containing type when trying to overload the mathematical operators so that they convert between different units. So, I have something like this: public interface ScalarUnit { } public class Duration : ScalarUnit { } public struct Scalar<T> where T : ScalarUnit { public readonly double Value; public Scalar(double Value) { this.Value = Value; } public static implicit operator double(Scalar<T> Value) { return Value.Value; } } public interface VectorUnit { } public class Displacement : VectorUnit { } public class Velocity : VectorUnit { } public struct Vector<T> where T : VectorUnit { #... public static Vector<Velocity> operator /(Vector<Displacement> v1, Scalar<Duration> v2) { return new Vector<Velocity>(v1.Magnitude / v2, v1.Direction); } } There aren't any errors for the + and - operators, where I'm just working on a Vector<T>, but when I substitute a unit for T, suddenly it doesn't like it. Is there a way to make this work? I figured it would work, since Displacement implements the VectorUnit interface, and I have where T : VectorUnit in the struct header. Am I at least on the right track here? I'm new to C# so I have difficulty understanding what's going on sometimes.

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  • Conversion constructor vs. conversion operator: precedence

    - by GRB
    Reading some questions here on SO about conversion operators and constructors got me thinking about the interaction between them, namely when there is an 'ambiguous' call. Consider the following code: class A; class B { public: B(){} B(const A&) //conversion constructor { cout << "called B's conversion constructor" << endl; } }; class A { public: operator B() //conversion operator { cout << "called A's conversion operator" << endl; return B(); } }; int main() { B b = A(); //what should be called here? apparently, A::operator B() return 0; } The above code displays "called A's conversion operator", meaning that the conversion operator is called as opposed to the constructor. If you remove/comment out the operator B() code from A, the compiler will happily switch over to using the constructor instead (with no other changes to the code). My questions are: Since the compiler doesn't consider B b = A(); to be an ambiguous call, there must be some type of precedence at work here. Where exactly is this precedence established? (a reference/quote from the C++ standard would be appreciated) From an object-oriented philosophical standpoint, is this the way the code should behave? Who knows more about how an A object should become a B object, A or B? According to C++, the answer is A -- is there anything in object-oriented practice that suggests this should be the case? To me personally, it would make sense either way, so I'm interested to know how the choice was made. Thanks in advance

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  • Dilemma with two types and operator +

    - by user35443
    I have small problem with operators. I have this code: public class A { public string Name { get; set; } public A() { } public A(string Name) { this.Name = Name; } public static implicit operator B(A a) { return new B(a.Name); } public static A operator+(A a, A b) { return new A(a.Name + " " + b.Name); } } public class B { public string Name { get; set; } public B() { } public B(string Name) { this.Name = Name; } public static implicit operator A(B b) { return new A(b.Name); } public static B operator +(B b, B a) { return new B(b.Name + " " + a.Name); } } Now I want to know, which's conversion operator will be called and which's addition operator will be called in this operation: new A("a") + new B("b"); Will it be operator of A, or of B? (Or both?) Thanks....

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  • How to modify a given class to use const operators

    - by zsero
    I am trying to solve my question regarding using push_back in more than one level. From the comments/answers it is clear that I have to: Create a copy operator which takes a const argument Modify all my operators to const But because this header file is given to me there is an operator what I cannot make into const. It is a simple: float & operator [] (int i) { return _item[i]; } In the given program, this operator is used to get and set data. My problem is that because I need to have this operator in the header file, I cannot turn all the other operators to const, what means I cannot insert a copy operator. How can I make all my operators into const, while preserving the functionality of the already written program? Here is the full declaration of the class: class Vector3f { float _item[3]; public: float & operator [] (int i) { return _item[i]; } Vector3f(float x, float y, float z) { _item[0] = x ; _item[1] = y ; _item[2] = z; }; Vector3f() {}; Vector3f & operator = ( const Vector3f& obj) { _item[0] = obj[0]; _item[1] = obj[1]; _item[2] = obj[2]; return *this; }; Vector3f & operator += ( const Vector3f & obj) { _item[0] += obj[0]; _item[1] += obj[1]; _item[2] += obj[2]; return *this; }; bool operator ==( const Vector3f & obj) { bool x = (_item[0] == obj[0]) && (_item[1] == obj[1]) && (_item[2] == obj[2]); return x; } // my copy operator Vector3f(const Vector3f& obj) { _item[0] += obj[0]; _item[1] += obj[1]; _item[2] += obj[2]; return this; } };

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  • DIVIDE vs division operator in #dax

    - by Marco Russo (SQLBI)
    Alberto Ferrari wrote an interesting article about DIVIDE performance in DAX. This new function has been introduced in SQL Server Analysis Services 2012 SP1, so it is available also in Excel 2013 (which still doesn’t have other features/fixes introduced by following Cumulative Updates…). The idea that instead of writing: IF ( Sales[Quantity] <> 0, Sales[Amount] / Sales[Quantity], BLANK () ) you can write: DIVIDE ( Sales[Amount], Sales[Quantity] ) There is a third optional argument in DIVIDE that defines the result in case the denominator (second argument) is zero, and by default its value is BLANK, so I omitted the third argument in my example. Using DIVIDE is very important, especially when you use a measure in MDX (for example in an Excel PivotTable) because it raise the chance that the non empty evaluation for the result is evaluated in bulk mode instead of cell-by-cell. However, from a DAX point of view, you might find it’s better to use the standard division operator removing the IF statement. I suggest you to read Alberto’s article, because you will find that an expression applying a filter using FILTER is faster than using CALCULATE, which is against any rule of thumb you might have read until now! Again, this is not always true, and depends on many conditions – trying to simplify, we might say that for a simple calculation, the query plan generated by FILTER could be more efficient – but, as usual, it depends, and 90% of the times using FILTER instead of CALCULATE produces slower performance. Do not take anything for granted, and always check the query plan when performance are your first issue!

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  • SQL SERVER – Fix: Error: 8117: Operand data type bit is invalid for sum operator

    - by pinaldave
    Here is the very interesting error I received from a reader. He has very interesting question. He attempted to use BIT filed in the SUM aggregation function and he got following error. He went ahead with various different datatype (i.e. INT, TINYINT etc) and he was able to do the SUM but with BIT he faced the problem. Error Received: Msg 8117, Level 16, State 1, Line 1 Operand data type bit is invalid for sum operator. Reproduction of the error: Set up the environment USE tempdb GO -- Preparing Sample Data CREATE TABLE TestTable (ID INT, Flag BIT) GO INSERT INTO TestTable (ID, Flag) SELECT 1, 0 UNION ALL SELECT 2, 1 UNION ALL SELECT 3, 0 UNION ALL SELECT 4, 1 GO SELECT * FROM TestTable GO Following script will work fine: -- This will work fine SELECT SUM(ID) FROM TestTable GO However following generate error: -- This will generate error SELECT SUM(Flag) FROM TestTable GO The workaround is to convert or cast the BIT to INT: -- Workaround of error SELECT SUM(CONVERT(INT, Flag)) FROM TestTable GO Clean up the setup -- Clean up DROP TABLE TestTable GO Workaround: As mentioned in above script the workaround is to covert the bit datatype to another friendly data types like INT, TINYINT etc. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Implicit conversion while using += operator?

    - by bdhar
    Conside the following code: int main() { signed char a = 10; a += a; // Line 5 a = a + a; return 0; } I am getting this warning at Line 5: d:\codes\operator cast\operator cast\test.cpp(5) : warning C4244: '+=' : conversion from 'int' to 'signed char', possible loss of data Does this mean that += operator makes an implicit cast of the right hand operator to int? P.S: I am using Visual studio 2005

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  • Returning recursive ternary freaks out

    - by David Titarenco
    Hi, assume this following function: int binaryTree::findHeight(node *n) { if (n == NULL) { return 0; } else { return 1 + max(findHeight(n->left), findHeight(n->right)); } } Pretty standard recursive treeHeight function for a given binary search tree binaryTree. Now, I was helping a friend (he's taking an algorithms course), and I ran into some weird issue with this function that I couldn't 100% explain to him. With max being defined as max(a,b) ((a)>(b)?(a):(b)) (which happens to be the max definition in windef.h), the recursive function freaks out (it runs something like n^n times where n is the tree height). This obviously makes checking the height of a tree with 3000 elements take very, very long. However, if max is defined via templating, like std does it, everything is okay. So using std::max fixed his problem. I just want to know why. Also, why does the countLeaves function work fine, using the same programmatic recursion? int binaryTree::countLeaves(node *n) { if (n == NULL) { return 0; } else if (n->left == NULL && n->right == NULL) { return 1; } else { return countLeaves(n->left) + countLeaves(n->right); } } Is it because in returning the ternary function, the values a => countLeaves(n->left) and b => countLeaves(n->right) were recursively double called simply because they were the resultants? Thank you!

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  • Ternary operator

    - by Antoine Leclair
    In PHP, I often use the ternary operator to add an attribute to an html element if it applies to the element in question. For example: <select name="blah"> <option value="1"<?= $blah == 1 ? ' selected="selected"' : '' ?>> One </option> <option value="2"<?= $blah == 2 ? ' selected="selected"' : '' ?>> Two </option> </select> I'm starting a project with Pylons using Mako for the templating. How can I achieve something similar? Right now, I see two possibilities that are not ideal. Solution 1: <select name="blah"> % if blah == 1: <option value="1" selected="selected">One</option> % else: <option value="1">One</option> % endif % if blah == 2: <option value="2" selected="selected">Two</option> % else: <option value="2">Two</option> % endif </select> Solution 2: <select name="blah"> <option value="1" % if blah == 1: selected="selected" % endif >One</option> <option value="2" % if blah == 2: selected="selected" % endif >Two</option> </select> In this particular case, the value is equal to the variable tested (value="1" = blah == 1), but I use the same pattern in other situations, like <?= isset($variable) ? ' value="$variable" : '' ?>. I am looking for a clean way to achieve this using Mako.

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  • converting if else statement to ternary

    - by EquinoX
    I have translated the following code using ternary. However, I knew there was something wrong with it. Can someone please point me into the right direction? ForwardA = 0; ForwardB = 0; //EX Hazard if (EXMEMRegWrite == 1) begin if (EXMEMrd != 0) if (EXMEMrd == IDEXrs) ForwardA = 2'b10; if (EXMEMrd == IDEXrt && IDEXTest == 0) ForwardB = 2'b10; end //MEM Hazard if (MEMWBRegWrite == 1) begin if (MEMWBrd != 0) begin if (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrs))) if (MEMWBrd == IDEXrs) ForwardA = 2'b01; if (IDEXTest == 0) begin if (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrt))) if (MEMWBrd == IDEXrt) ForwardB = 2'b01; end end end ForwardA = (MEMWBRegWrite && MEMWBrd != 0 && (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrs))) && (MEMWBrd == IDEXrs)) ? 2'b01 : ((EXMEMRegWrite && EXMEMrd != 0 && EXMEMrd == IDEXrs) ? 2'b10 : 0); ForwardB = (IDEXTest == 0 && MEMWBRegWrite && MEMWBrd != 0 && (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrt))) && (MEMWBrd == IDEXrs)) ? 2'b01 : ((EXMEMRegWrite && EXMEMrd != 0 && EXMEMrd == IDEXrt && IDEXTest == 0) ? 2'b10 : 0);

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