Search Results

Search found 5377 results on 216 pages for 'explicit cast operator'.

Page 2/216 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • conditional operator in C question

    - by Seephor
    I just have a quick question about the conditional operator. Still a budding programmer here. I am given x = 1, y = 2, and z = 3. I want to know, why after this statement: y += x-- ? z++ : --z; That y is 5. The values after the statement are x = 0, y = 5, and z = 4. I know the way the conditional operator works is that it is formatted like this: variable = condition ? value if true : value if false. For the condition, y += x-- , how does y become 5? I can only see 2 (2 += 0) and 3 (2 += 1)(then x-- becomes zero) as possibilities. Any help is much appreciated. :)

    Read the article

  • Operator Overloading in C++ as int + obj

    - by Azher
    Hi Guys, I have following class:- class myclass { size_t st; myclass(size_t pst) { st=pst; } operator int() { return (int)st; } int operator+(int intojb) { return int(st) + intobj; } }; this works fine as long as I use it like this:- char* src="This is test string"; int i= myclass(strlen(src)) + 100; but I am unable to do this:- int i= 100+ myclass(strlen(src)); Any idea, how can I achieve this?? Thanks in advance. Regards,

    Read the article

  • What do you think about ??= operator in C#?

    - by TN
    Do you think that C# will support something like ??= operator? Instead of this: if (list == null) list = new List<int>(); It might be possible to write: list ??= new List<int>(); Now, I could use (but it seems to me not well readable): list = list ?? new List<int>();

    Read the article

  • Java conditional operator ?: result type

    - by Wangnick
    I'm a bit puzzled about the conditional operator. Consider the following two lines: Float f1 = false? 1.0f: null; Float f2 = false? 1.0f: false? 1.0f: null; Why does f1 become null and the second statement throws a NullPointerException? Langspec-3.0 para 15.25 sais: Otherwise, the second and third operands are of types S1 and S2 respectively. Let T1 be the type that results from applying boxing conversion to S1, and let T2 be the type that results from applying boxing conversion to S2. The type of the conditional expression is the result of applying capture conversion (§5.1.10) to lub(T1, T2) (§15.12.2.7). So for false?1.0f:null T1 is Float and T2 is the null type. But what is the result of lub(T1,T2)? This para 15.12.2.7 is just a bit too much ... BTW, I'm using 1.6.0_18 on Windows. PS: I know that Float f2 = false? (Float) 1.0f: false? (Float) 1.0f: null; doesn't throw NPE.

    Read the article

  • Ternary operator in VB.NET

    - by Jalpesh P. Vadgama
    We all know about Ternary operator in C#.NET. I am a big fan of ternary operator and I like to use it instead of using IF..Else. Those who don’t know about ternary operator please go through below link. http://msdn.microsoft.com/en-us/library/ty67wk28(v=vs.80).aspx Here you can see ternary operator returns one of the two values based on the condition. See following example. bool value = false;string output=string.Empty;//using If conditionif (value==true) output ="True";else output="False";//using tenary operatoroutput = value == true ? "True" : "False"; In the above example you can see how we produce same output with the ternary operator without using If..Else statement. Recently in one of the project I was working with VB.NET language and I was eager to know if there is a ternary operator equivalent there or not. After searching on internet I have found two ways to do it. IF operator which works for VB.NET 2008 and higher version and IIF operator which is there since VB 6.0. So let’s check same above example with both of this operators. So let’s create a console application which has following code. Module Module1 Sub Main() Dim value As Boolean = False Dim output As String = String.Empty ''Output using if else statement If value = True Then output = "True" Else output = "False" Console.WriteLine("Output Using If Loop") Console.WriteLine(output) output = If(value = True, "True", "False") Console.WriteLine("Output using If operator") Console.WriteLine(output) output = IIf(value = True, "True", "False") Console.WriteLine("Output using IIF Operator") Console.WriteLine(output) Console.ReadKey() End If End SubEnd Module As you can see in the above code I have written all three-way to condition check using If.Else statement and If operator and IIf operator. You can see that both IIF and If operator has three parameter first parameter is the condition which you need to check and then another parameter is true part of you need to put thing which you need as output when condition is ‘true’. Same way third parameter is for the false part where you need to put things which you need as output when condition as ‘false’. Now let’s run that application and following is the output as expected. That’s it. You can see all three ways are producing same output. Hope you like it. Stay tuned for more..Till then Happy Programming.

    Read the article

  • Purpose of Explicit Default Constructors

    - by Dennis Zickefoose
    I recently noticed a class in C++0x that calls for an explicit default constructor. However, I'm failing to come up with a scenario in which a default constructor can be called implicitly. It seems like a rather pointless specifier. I thought maybe it would disallow Class c; in favor of Class c = Class(); but that does not appear to be the case. Some relevant quotes from the C++0x FCD, since it is easier for me to navigate [similar text exists in C++03, if not in the same places] 12.3.1.3 [class.conv.ctor] A default constructor may be an explicit constructor; such a constructor will be used to perform default-initialization or value initialization (8.5). It goes on to provide an example of an explicit default constructor, but it simply mimics the example I provided above. 8.5.6 [decl.init] To default-initialize an object of type T means: — if T is a (possibly cv-qualified) class type (Clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor); 8.5.7 [decl.init] To value-initialize an object of type T means: — if T is a (possibly cv-qualified) class type (Clause 9) with a user-provided constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor); In both cases, the standard calls for the default constructor to be called. But that is what would happen if the default constructor were non-explicit. For completeness sake: 8.5.11 [decl.init] If no initializer is specified for an object, the object is default-initialized; From what I can tell, this just leaves conversion from no data. Which doesn't make sense. The best I can come up with would be the following: void function(Class c); int main() { function(); //implicitly convert from no parameter to a single parameter } But obviously that isn't the way C++ handles default arguments. What else is there that would make explicit Class(); behave differently from Class();? The specific example that generated this question was std::function [20.8.14.2 func.wrap.func]. It requires several converting constructors, none of which are marked explicit, but the default constructor is.

    Read the article

  • IEnumerable.Cast not calling cast overload

    - by Martin Neal
    I'm not understanding something about the way .Cast works. I have an explicit (though implicit also fails) cast defined which seems to work when I use it "regularly", but not when I try to use .Cast. Why? Here is some compilable code that demonstrates my problem. public class Class1 { public string prop1 { get; set; } public int prop2 { get; set; } public static explicit operator Class2(Class1 c1) { return new Class2() { prop1 = c1.prop1, prop2 = c1.prop2 }; } } public class Class2 { public string prop1 { get; set; } public int prop2 { get; set; } } void Main() { Class1[] c1 = new Class1[] { new Class1() {prop1 = "asdf",prop2 = 1}}; //works Class2 c2 = (Class2)c1[0]; //doesn't work: Compiles, but throws at run-time //InvalidCastException: Unable to cast object of type 'Class1' to type 'Class2'. Class2 c3 = c1.Cast<Class2>().First(); }

    Read the article

  • C++ cast syntax styles

    - by palm3D
    A question related to Regular cast vs. static_cast vs. dynamic_cast: What cast syntax style do you prefer in C++? C-style cast syntax: (int)foo C++-style cast syntax: static_cast<int>(foo) constructor syntax: int(foo) They may not translate to exactly the same instructions (do they?) but their effect should be the same (right?). If you're just casting between the built-in numeric types, I find C++-style cast syntax too verbose. As a former Java coder I tend to use C-style cast syntax instead, but my local C++ guru insists on using constructor syntax. What do you think?

    Read the article

  • Plan Operator Tuesday round-up

    - by Rob Farley
    Eighteen posts for T-SQL Tuesday #43 this month, discussing Plan Operators. I put them together and made the following clickable plan. It’s 1000px wide, so I hope you have a monitor wide enough. Let me explain this plan for you (people’s names are the links to the articles on their blogs – the same links as in the plan above). It was clearly a SELECT statement. Wayne Sheffield (@dbawayne) wrote about that, so we start with a SELECT physical operator, leveraging the logical operator Wayne Sheffield. The SELECT operator calls the Paul White operator, discussed by Jason Brimhall (@sqlrnnr) in his post. The Paul White operator is quite remarkable, and can consume three streams of data. Let’s look at those streams. The first pulls data from a Table Scan – Boris Hristov (@borishristov)’s post – using parallel threads (Bradley Ball – @sqlballs) that pull the data eagerly through a Table Spool (Oliver Asmus – @oliverasmus). A scalar operation is also performed on it, thanks to Jeffrey Verheul (@devjef)’s Compute Scalar operator. The second stream of data applies Evil (I figured that must mean a procedural TVF, but could’ve been anything), courtesy of Jason Strate (@stratesql). It performs this Evil on the merging of parallel streams (Steve Jones – @way0utwest), which suck data out of a Switch (Paul White – @sql_kiwi). This Switch operator is consuming data from up to four lookups, thanks to Kalen Delaney (@sqlqueen), Rick Krueger (@dataogre), Mickey Stuewe (@sqlmickey) and Kathi Kellenberger (@auntkathi). Unfortunately Kathi’s name is a bit long and has been truncated, just like in real plans. The last stream performs a join of two others via a Nested Loop (Matan Yungman – @matanyungman). One pulls data from a Spool (my post – @rob_farley) populated from a Table Scan (Jon Morisi). The other applies a catchall operator (the catchall is because Tamera Clark (@tameraclark) didn’t specify any particular operator, and a catchall is what gets shown when SSMS doesn’t know what to show. Surprisingly, it’s showing the yellow one, which is about cursors. Hopefully that’s not what Tamera planned, but anyway...) to the output from an Index Seek operator (Sebastian Meine – @sqlity). Lastly, I think everyone put in 110% effort, so that’s what all the operators cost. That didn’t leave anything for me, unfortunately, but that’s okay. Also, because he decided to use the Paul White operator, Jason Brimhall gets 0%, and his 110% was given to Paul’s Switch operator post. I hope you’ve enjoyed this T-SQL Tuesday, and have learned something extra about Plan Operators. Keep your eye out for next month’s one by watching the Twitter Hashtag #tsql2sday, and why not contribute a post to the party? Big thanks to Adam Machanic as usual for starting all this. @rob_farley

    Read the article

  • operator overloading of stream extraction operator in C++ help

    - by Crystal
    I'm having some trouble overloading my stream extraction operator in C++ for a hw assignment. I'm not really sure why I am getting these compile errors since I thought I was doing it right... Here is my code: Complex.h #ifndef COMPLEX_H #define COMPLEX_H class Complex { //friend ostream &operator<<(ostream &output, const Complex &complexObj) const; public: Complex(double = 0.0, double = 0.0); // constructor Complex operator+(const Complex &) const; // addition Complex operator-(const Complex &) const; // subtraction void print() const; // output private: double real; // real part double imaginary; // imaginary part }; #endif Complex.cpp #include <iostream> #include "Complex.h" using namespace std; // Constructor Complex::Complex(double realPart, double imaginaryPart) : real(realPart), imaginary(imaginaryPart) { } // addition operator Complex Complex::operator+(const Complex &operand2) const { return Complex(real + operand2.real, imaginary + operand2.imaginary); } // subtraction operator Complex Complex::operator-(const Complex &operand2) const { return Complex(real - operand2.real, imaginary - operand2.imaginary); } // Overload << operator ostream &Complex::operator<<(ostream &output, const Complex &complexObj) const { cout << '(' << complexObj.real << ", " << complexObj.imaginary << ')'; return output; // returning output allows chaining } // display a Complex object in the form: (a, b) void Complex::print() const { cout << '(' << real << ", " << imaginary << ')'; } main.cpp #include <iostream> #include "Complex.h" using namespace std; int main() { Complex x; Complex y(4.3, 8.2); Complex z(3.3, 1.1); cout << "x: "; x.print(); cout << "\ny: "; y.print(); cout << "\nz: "; z.print(); x = y + z; cout << "\n\nx = y + z: " << endl; x.print(); cout << " = "; y.print(); cout << " + "; z.print(); x = y - z; cout << "\n\nx = y - z: " << endl; x.print(); cout << " = "; y.print(); cout << " - "; z.print(); cout << endl; } Compile erros: complex.cpp(23) : error C2039: '<<' : is not a member of 'Complex' complex.h(5) : see declaration of 'Complex' complex.cpp(24) : error C2270: '<<' : modifiers not allowed on nonmember functions complex.cpp(25) : error C2248: 'Complex::real' : cannot access private member declared in class 'Complex' complex.h(13) : see declaration of 'Complex::real' complex.h(5) : see declaration of 'Complex' complex.cpp(25) : error C2248: 'Complex::imaginary' : cannot access private member declared in class 'Complex' complex.h(14) : see declaration of 'Complex::imaginary' complex.h(5) : see declaration of 'Complex' Thanks!

    Read the article

  • C++ Operator Ambiguity

    - by Scott
    Forgive me, for I am fairly new to C++, but I am having some trouble regarding operator ambiguity. I think it is compiler-specific, for the code compiled on my desktop. However, it fails to compile on my laptop. I think I know what's going wrong, but I don't see an elegant way around it. Please let me know if I am making an obvious mistake. Anyhow, here's what I'm trying to do: I have made my own vector class called Vector4 which looks something like this: class Vector4 { private: GLfloat vector[4]; ... } Then I have these operators, which are causing the problem: operator GLfloat* () { return vector; } operator const GLfloat* () const { return vector; } GLfloat& operator [] (const size_t i) { return vector[i]; } const GLfloat& operator [] (const size_t i) const { return vector[i]; } I have the conversion operator so that I can pass an instance of my Vector4 class to glVertex3fv, and I have subscripting for obvious reasons. However, calls that involve subscripting the Vector4 become ambiguous to the compiler: enum {x, y, z, w} Vector4 v(1.0, 2.0, 3.0, 4.0); glTranslatef(v[x], v[y], v[z]); Here are the candidates: candidate 1: const GLfloat& Vector4:: operator[](size_t) const candidate 2: operator[](const GLfloat*, int) <built-in> Why would it try to convert my Vector4 to a GLfloat* first when the subscript operator is already defined on Vector4? Is there a simple way around this that doesn't involve typecasting? Am I just making a silly mistake? Thanks for any help in advance.

    Read the article

  • Why does this cast to Base class in virtual function give a segmentation fault?

    - by dehmann
    I want to print out a derived class using the operator<<. When I print the derived class, I want to first print its base and then its own content. But I ran into some trouble (see segfault below): class Base { public: friend std::ostream& operator<<(std::ostream&, const Base&); virtual void Print(std::ostream& out) const { out << "BASE!"; } }; std::ostream& operator<<(std::ostream& out, const Base& b) { b.Print(out); return out; } class Derived : public Base { public: virtual void Print(std::ostream& out) const { out << "My base: "; //((const Base*)this)->Print(out); // infinite, calls this fct recursively //((Base*)this)->Print(out); // segfault (from infinite loop?) ((Base)*this).Print(out); // OK out << " ... and myself."; } }; int main(int argc, char** argv){ Derived d; std::cout << d; return 0; } Why can't I cast in one of these ways? ((const Base*)this)->Print(out); // infinite, calls this fct recursively ((Base*)this)->Print(out); // segfault (from infinite loop?)

    Read the article

  • ASP.NET MVC 3: Implicit and Explicit code nuggets with Razor

    - by ScottGu
    This is another in a series of posts I’m doing that cover some of the new ASP.NET MVC 3 features: New @model keyword in Razor (Oct 19th) Layouts with Razor (Oct 22nd) Server-Side Comments with Razor (Nov 12th) Razor’s @: and <text> syntax (Dec 15th) Implicit and Explicit code nuggets with Razor (today) In today’s post I’m going to discuss how Razor enables you to both implicitly and explicitly define code nuggets within your view templates, and walkthrough some code examples of each of them.  Fluid Coding with Razor ASP.NET MVC 3 ships with a new view-engine option called “Razor” (in addition to the existing .aspx view engine).  You can learn more about Razor, why we are introducing it, and the syntax it supports from my Introducing Razor blog post. Razor minimizes the number of characters and keystrokes required when writing a view template, and enables a fast, fluid coding workflow. Unlike most template syntaxes, you do not need to interrupt your coding to explicitly denote the start and end of server blocks within your HTML. The Razor parser is smart enough to infer this from your code. This enables a compact and expressive syntax which is clean, fast and fun to type. For example, the Razor snippet below can be used to iterate a collection of products and output a <ul> list of product names that link to their corresponding product pages: When run, the above code generates output like below: Notice above how we were able to embed two code nuggets within the content of the foreach loop.  One of them outputs the name of the Product, and the other embeds the ProductID within a hyperlink.  Notice that we didn’t have to explicitly wrap these code-nuggets - Razor was instead smart enough to implicitly identify where the code began and ended in both of these situations.  How Razor Enables Implicit Code Nuggets Razor does not define its own language.  Instead, the code you write within Razor code nuggets is standard C# or VB.  This allows you to re-use your existing language skills, and avoid having to learn a customized language grammar. The Razor parser has smarts built into it so that whenever possible you do not need to explicitly mark the end of C#/VB code nuggets you write.  This makes coding more fluid and productive, and enables a nice, clean, concise template syntax.  Below are a few scenarios that Razor supports where you can avoid having to explicitly mark the beginning/end of a code nugget, and instead have Razor implicitly identify the code nugget scope for you: Property Access Razor allows you to output a variable value, or a sub-property on a variable that is referenced via “dot” notation: You can also use “dot” notation to access sub-properties multiple levels deep: Array/Collection Indexing: Razor allows you to index into collections or arrays: Calling Methods: Razor also allows you to invoke methods: Notice how for all of the scenarios above how we did not have to explicitly end the code nugget.  Razor was able to implicitly identify the end of the code block for us. Razor’s Parsing Algorithm for Code Nuggets The below algorithm captures the core parsing logic we use to support “@” expressions within Razor, and to enable the implicit code nugget scenarios above: Parse an identifier - As soon as we see a character that isn't valid in a C# or VB identifier, we stop and move to step 2 Check for brackets - If we see "(" or "[", go to step 2.1., otherwise, go to step 3  Parse until the matching ")" or "]" (we track nested "()" and "[]" pairs and ignore "()[]" we see in strings or comments) Go back to step 2 Check for a "." - If we see one, go to step 3.1, otherwise, DO NOT ACCEPT THE "." as code, and go to step 4 If the character AFTER the "." is a valid identifier, accept the "." and go back to step 1, otherwise, go to step 4 Done! Differentiating between code and content Step 3.1 is a particularly interesting part of the above algorithm, and enables Razor to differentiate between scenarios where an identifier is being used as part of the code statement, and when it should instead be treated as static content: Notice how in the snippet above we have ? and ! characters at the end of our code nuggets.  These are both legal C# identifiers – but Razor is able to implicitly identify that they should be treated as static string content as opposed to being part of the code expression because there is whitespace after them.  This is pretty cool and saves us keystrokes. Explicit Code Nuggets in Razor Razor is smart enough to implicitly identify a lot of code nugget scenarios.  But there are still times when you want/need to be more explicit in how you scope the code nugget expression.  The @(expression) syntax allows you to do this: You can write any C#/VB code statement you want within the @() syntax.  Razor will treat the wrapping () characters as the explicit scope of the code nugget statement.  Below are a few scenarios where we could use the explicit code nugget feature: Perform Arithmetic Calculation/Modification: You can perform arithmetic calculations within an explicit code nugget: Appending Text to a Code Expression Result: You can use the explicit expression syntax to append static text at the end of a code nugget without having to worry about it being incorrectly parsed as code: Above we have embedded a code nugget within an <img> element’s src attribute.  It allows us to link to images with URLs like “/Images/Beverages.jpg”.  Without the explicit parenthesis, Razor would have looked for a “.jpg” property on the CategoryName (and raised an error).  By being explicit we can clearly denote where the code ends and the text begins. Using Generics and Lambdas Explicit expressions also allow us to use generic types and generic methods within code expressions – and enable us to avoid the <> characters in generics from being ambiguous with tag elements. One More Thing….Intellisense within Attributes We have used code nuggets within HTML attributes in several of the examples above.  One nice feature supported by the Razor code editor within Visual Studio is the ability to still get VB/C# intellisense when doing this. Below is an example of C# code intellisense when using an implicit code nugget within an <a> href=”” attribute: Below is an example of C# code intellisense when using an explicit code nugget embedded in the middle of a <img> src=”” attribute: Notice how we are getting full code intellisense for both scenarios – despite the fact that the code expression is embedded within an HTML attribute (something the existing .aspx code editor doesn’t support).  This makes writing code even easier, and ensures that you can take advantage of intellisense everywhere. Summary Razor enables a clean and concise templating syntax that enables a very fluid coding workflow.  Razor’s ability to implicitly scope code nuggets reduces the amount of typing you need to perform, and leaves you with really clean code. When necessary, you can also explicitly scope code expressions using a @(expression) syntax to provide greater clarity around your intent, as well as to disambiguate code statements from static markup. Hope this helps, Scott P.S. In addition to blogging, I am also now using Twitter for quick updates and to share links. Follow me at: twitter.com/scottgu

    Read the article

  • wrong operator() overload called

    - by user313202
    okay, I am writing a matrix class and have overloaded the function call operator twice. The core of the matrix is a 2D double array. I am using the MinGW GCC compiler called from a windows console. the first overload is meant to return a double from the array (for viewing an element). the second overload is meant to return a reference to a location in the array (for changing the data in that location. double operator()(int row, int col) const ; //allows view of element double &operator()(int row, int col); //allows assignment of element I am writing a testing routine and have discovered that the "viewing" overload never gets called. for some reason the compiler "defaults" to calling the overload that returns a reference when the following printf() statement is used. fprintf(outp, "%6.2f\t", testMatD(i,j)); I understand that I'm insulting the gods by writing my own matrix class without using vectors and testing with C I/O functions. I will be punished thoroughly in the afterlife, no need to do it here. Ultimately I'd like to know what is going on here and how to fix it. I'd prefer to use the cleaner looking operator overloads rather than member functions. Any ideas? -Cal the matrix class: irrelevant code omitted class Matrix { public: double getElement(int row, int col)const; //returns the element at row,col //operator overloads double operator()(int row, int col) const ; //allows view of element double &operator()(int row, int col); //allows assignment of element private: //data members double **array; //pointer to data array }; double Matrix::getElement(int row, int col)const{ //transform indices into true coordinates (from sorted coordinates //only row needs to be transformed (user can only sort by row) row = sortedArray[row]; result = array[usrZeroRow+row][usrZeroCol+col]; return result; } //operator overloads double Matrix::operator()(int row, int col) const { //this overload is used when viewing an element return getElement(row,col); } double &Matrix::operator()(int row, int col){ //this overload is used when placing an element return array[row+usrZeroRow][col+usrZeroCol]; } The testing program: irrelevant code omitted int main(void){ FILE *outp; outp = fopen("test_output.txt", "w+"); Matrix testMatD(5,7); //construct 5x7 matrix //some initializations omitted fprintf(outp, "%6.2f\t", testMatD(i,j)); //calls the wrong overload }

    Read the article

  • Operator Overloading in C

    - by Leif Andersen
    In C++, I can change the operator on a specific class by doing something like this: MyClass::operator==/*Or some other operator such as =, >, etc.*/(Const MyClass rhs) { /* Do Stuff*/; } But with there being no classes (built in by default) in C. So, how could I do operator overloading for just general functions? For example, if I remember correctly, importing stdlib.h gives you the - operator, which is just syntactic sugar for (*strcut_name).struct_element. So how can I do this in C? Thank you.

    Read the article

  • Explain this C++ operator definition

    - by David Johnstone
    I have the following operator defined in a C++ class called StringProxy: operator std::string&() { return m_string; } a) What is this and how does this work? I understand the idea of operator overloading, but they normally look like X operator+(double i). b) Given an instance of StringProxy, how can I use this operator to get the m_string?

    Read the article

  • Why can't I use interface with explicit operator?

    - by theburningmonk
    Hi, I'm just wondering if anyone knows the reason why you are not allowed to use interfaces with the implicit or explicit operators? E.g. this raises compile time error: public static explicit operator MyPlayer(IPlayer player) { ... } "user-defined conversions to or from an interface are not allowed" Thanks,

    Read the article

  • C++ Operator overloading - 'recreating the Vector'

    - by Wallter
    I am currently in a collage second level programing course... We are working on operator overloading... to do this we are to rebuild the vector class... I was building the class and found that most of it is based on the [] operator. When I was trying to implement the + operator I run into a weird error that my professor has not seen before (apparently since the class switched IDE's from MinGW to VS express...) (I am using Visual Studio Express 2008 C++ edition...) Vector.h #include <string> #include <iostream> using namespace std; #ifndef _VECTOR_H #define _VECTOR_H const int DEFAULT_VECTOR_SIZE = 5; class Vector { private: int * data; int size; int comp; public: inline Vector (int Comp = 5,int Size = 0) : comp(Comp), size(Size) { if (comp > 0) { data = new int [comp]; } else { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; } } int size_ () const { return size; } int comp_ () const { return comp; } bool push_back (int); bool push_front (int); void expand (); void expand (int); void clear (); const string at (int); int operator[ ](int); Vector& operator+ (Vector&); Vector& operator- (const Vector&); bool operator== (const Vector&); bool operator!= (const Vector&); ~Vector() { delete [] data; } }; ostream& operator<< (ostream&, const Vector&); #endif Vector.cpp #include <iostream> #include <string> #include "Vector.h" using namespace std; const string Vector::at(int i) { this[i]; } void Vector::expand() { expand(size); } void Vector::expand(int n ) { int * newdata = new int [comp * 2]; if (*data != NULL) { for (int i = 0; i <= (comp); i++) { newdata[i] = data[i]; } newdata -= comp; comp += n; delete [] data; *data = *newdata; } else if ( *data == NULL || comp == 0) { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; size = 0; } } bool Vector::push_back(int n) { if (comp = 0) { expand(); } for (int k = 0; k != 2; k++) { if ( size != comp ){ data[size] = n; size++; return true; } else { expand(); } } return false; } void Vector::clear() { delete [] data; comp = 0; size = 0; } int Vector::operator[] (int place) { return (data[place]); } Vector& Vector::operator+ (Vector& n) { int temp_int = 0; if (size > n.size_() || size == n.size_()) { temp_int = size; } else if (size < n.size_()) { temp_int = n.size_(); } Vector newone(temp_int); int temp_2_int = 0; for ( int j = 0; j <= temp_int && j <= n.size_() && j <= size; j++) { temp_2_int = n[j] + data[j]; newone[j] = temp_2_int; } //////////////////////////////////////////////////////////// return newone; //////////////////////////////////////////////////////////// } ostream& operator<< (ostream& out, const Vector& n) { for (int i = 0; i <= n.size_(); i++) { //////////////////////////////////////////////////////////// out << n[i] << " "; //////////////////////////////////////////////////////////// } return out; } Errors: out << n[i] << " "; error C2678: binary '[' : no operator found which takes a left-hand operand of type 'const Vector' (or there is no acceptable conversion) return newone; error C2106: '=' : left operand must be l-value As stated above, I am a student going into Computer Science as my selected major I would appreciate tips, pointers, and better ways to do stuff :D

    Read the article

  • Are free operator->* overloads evil?

    - by Potatoswatter
    I was perusing section 13.5 after refuting the notion that built-in operators do not participate in overload resolution, and noticed that there is no section on operator->*. It is just a generic binary operator. Its brethren, operator->, operator*, and operator[], are all required to be non-static member functions. This precludes definition of a free function overload to an operator commonly used to obtain a reference from an object. But the uncommon operator->* is left out. In particular, operator[] has many similarities. It is binary (they missed a golden opportunity to make it n-ary), and it accepts some kind of container on the left and some kind of locator on the right. Its special-rules section, 13.5.5, doesn't seem to have any actual effect except to outlaw free functions. (And that restriction even precludes support for commutativity!) So, for example, this is perfectly legal (in C++0x, remove obvious stuff to translate to C++03): #include <utility> #include <iostream> #include <type_traits> using namespace std; template< class F, class S > typename common_type< F,S >::type operator->*( pair<F,S> const &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( pair<T,T> &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( bool l, pair<T,T> &r ) { return l? r.second : r.first; } int main() { auto x = make_pair( 1, 2.3 ); cerr << x->*false << " " << x->*4 << endl; auto y = make_pair( 5, 6 ); y->*(0) = 7; y->*0->*y = 8; // evaluates to 7->*y = y.second cerr << y.first << " " << y.second << endl; } I can certainly imagine myself giving into temp[la]tation. For example, scaled indexes for vector: v->*matrix_width[2][5] = x; Did the standards committee forget to prevent this, was it considered too ugly to bother, or are there real-world use cases?

    Read the article

  • Overloading *(iterator + n) and *(n + iterator) in a C++ iterator class?

    - by exscape
    (Note: I'm writing this project for learning only; comments about it being redundant are... uh, redundant. ;) I'm trying to implement a random access iterator, but I've found very little literature on the subject, so I'm going by trial and error combined with Wikpedias list of operator overload prototypes. It's worked well enough so far, but I've hit a snag. Code such as exscape::string::iterator i = string_instance.begin(); std::cout << *i << std::endl; works, and prints the first character of the string. However, *(i + 1) doesn't work, and neither does *(1 + i). My full implementation would obviously be a bit too much, but here's the gist of it: namespace exscape { class string { friend class iterator; ... public: class iterator : public std::iterator<std::random_access_iterator_tag, char> { ... char &operator*(void) { return *p; // After some bounds checking } char *operator->(void) { return p; } char &operator[](const int offset) { return *(p + offset); // After some bounds checking } iterator &operator+=(const int offset) { p += offset; return *this; } const iterator operator+(const int offset) { iterator out (*this); out += offset; return out; } }; }; } int main() { exscape::string s = "ABCDEF"; exscape::string::iterator i = s.begin(); std::cout << *(i + 2) << std::endl; } The above fails with (line 632 is, of course, the *(i + 2) line): string.cpp: In function ‘int main()’: string.cpp:632: error: no match for ‘operator*’ in ‘*exscape::string::iterator::operator+(int)(2)’ string.cpp:105: note: candidates are: char& exscape::string::iterator::operator*() *(2 + i) fails with: string.cpp: In function ‘int main()’: string.cpp:632: error: no match for ‘operator+’ in ‘2 + i’ string.cpp:434: note: candidates are: exscape::string exscape::operator+(const char*, const exscape::string&) My guess is that I need to do some more overloading, but I'm not sure what operator I'm missing.

    Read the article

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