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  • Custom string class (C++)

    - by Sanctus2099
    Hey guys. I'm trying to write my own C++ String class for educational and need purposes. The first thing is that I don't know that much about operators and that's why I want to learn them. I started writing my class but when I run it it blocks the program but does not do any crash. Take a look at the following code please before reading further: class CString { private: char* cstr; public: CString(); CString(char* str); CString(CString& str); ~CString(); operator char*(); operator const char*(); CString operator+(const CString& q)const; CString operator=(const CString& q); }; First of all I'm not so sure I declared everything right. I tried googleing about it but all the tutorials about overloading explain the basic ideea which is very simple but lack to explain how and when each thing is called. For instance in my = operator the program calls CString(CString& str); but I have no ideea why. I have also attached the cpp file below: CString::CString() { cstr=0; } CString::CString(char *str) { cstr=new char[strlen(str)]; strcpy(cstr,str); } CString::CString(CString& q) { if(this==&q) return; cstr = new char[strlen(q.cstr)+1]; strcpy(cstr,q.cstr); } CString::~CString() { if(cstr) delete[] cstr; } CString::operator char*() { return cstr; } CString::operator const char* () { return cstr; } CString CString::operator +(const CString &q) const { CString s; s.cstr = new char[strlen(cstr)+strlen(q.cstr)+1]; strcpy(s.cstr,cstr); strcat(s.cstr,q.cstr); return s; } CString CString::operator =(const CString &q) { if(this!=&q) { if(cstr) delete[] cstr; cstr = new char[strlen(q.cstr)+1]; strcpy(cstr,q.cstr); } return *this; } For testing I used a code just as simple as this CString a = CString("Hello") + CString(" World"); printf(a); I tried debugging it but at a point I get lost. First it calls the constructor 2 times for "hello" and for " world". Then it get's in the + operator which is fine. Then it calls the constructor for the empty string. After that it get's into "CString(CString& str)" and now I'm lost. Why is this happening? After this I noticed my string containing "Hello World" is in the destructor (a few times in a row). Again I'm very puzzeled. After converting again from char* to Cstring and back and forth it stops. It never get's into the = operator but neither does it go further. printf(a) is never reached. I use VisualStudio 2010 for this but it's basically just standard c++ code and thus I don't think it should make that much of a difference

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  • Why isn't my operator overloading working properly?

    - by Mithrax
    I have the following Polynomial class I'm working on: #include <iostream> using namespace std; class Polynomial { //define private member functions private: int coef[100]; // array of coefficients // coef[0] would hold all coefficients of x^0 // coef[1] would hold all x^1 // coef[n] = x^n ... int deg; // degree of polynomial (0 for the zero polynomial) //define public member functions public: Polynomial::Polynomial() //default constructor { for ( int i = 0; i < 100; i++ ) { coef[i] = 0; } } void set ( int a , int b ) //setter function { //coef = new Polynomial[b+1]; coef[b] = a; deg = degree(); } int degree() { int d = 0; for ( int i = 0; i < 100; i++ ) if ( coef[i] != 0 ) d = i; return d; } void print() { for ( int i = 99; i >= 0; i-- ) { if ( coef[i] != 0 ) { cout << coef[i] << "x^" << i << " "; } } } // use Horner's method to compute and return the polynomial evaluated at x int evaluate ( int x ) { int p = 0; for ( int i = deg; i >= 0; i-- ) p = coef[i] + ( x * p ); return p; } // differentiate this polynomial and return it Polynomial differentiate() { if ( deg == 0 ) { Polynomial t; t.set ( 0, 0 ); return t; } Polynomial deriv;// = new Polynomial ( 0, deg - 1 ); deriv.deg = deg - 1; for ( int i = 0; i < deg; i++ ) deriv.coef[i] = ( i + 1 ) * coef[i + 1]; return deriv; } Polynomial Polynomial::operator + ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator += ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator *= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator * ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); return c; } }; int main() { Polynomial a, b, c, d; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; c = a * b; // (7x^4 + x^2) * (6x^3 - 3x^2) c.print(); cout << "\n"; d = c.differentiate().differentiate(); d.print(); cout << "\n"; cout << c.evaluate ( 2 ); //substitue x with 2 cin.get(); } Now, I have the "-" operator overloaded and it works fine: Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } However, I'm having difficulty with my "-=" operator: Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); // overwrite value of 'a' with the newly computed 'c' before returning 'a' for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } I just slightly modified my "-" operator method to overwrite the value in 'a' and return 'a', and just use the 'c' polynomial as a temp. I've put in some debug print statement and I confirm that at the time of computation, both: c = a - b; and a -= b; are computed to the same value. However, when I go to print them, their results are different: Polynomial a, b; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; Result: 7x^4 -6x^3 4x^2 7x^4 1x^2 Why is my c = a - b and a -= b giving me different results when I go to print them?

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  • Conditional Operator Example

    - by mbcrump
    If you haven’t taken the time to learn conditional operators, then now is the time. I’ve added a quick and dirty example for those on the forums.   Code Snippet using System; using System.Net.Mail; using System.Net; using System.Globalization; using System.Windows.Forms;   class Demo {     //Please use conditional statements in your code. See example below.       public static void Main()     {         int dollars = 10;           //Bad Coder Bad !!! Don't do this         if (dollars == 1)         {             Console.WriteLine("Please deposit {0} dollar.", dollars);         }         else         {             Console.WriteLine("Please deposit {0} dollars.", dollars);         }             //Good Coder Good !!! Do this         Console.WriteLine("Please deposit {0} dollar{1}.", dollars, dollars == 1 ? ' ' : 's');         //                                                          expression   ? true : false           Console.ReadLine();          } }

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  • Showplan Operator of the Week – BookMark/Key Lookup

    Fabiano continues in his mission to describe the major Showplan Operators used by SQL Server's Query Optimiser. This week he meets a star, the Key Lookup, a stalwart performer, but most famous for its role in ill-performing queries where an index does not 'cover' the data required to execute the query. If you understand why, and in what circumstances, key lookups are slow, it helps greatly with optimising query performance.

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  • LINQ and the use of Repeat and Range operator

    - by vik20000in
    LINQ is also very useful when it comes to generation of range or repetition of data.  We can generate a range of data with the help of the range method.     var numbers =         from n in Enumerable.Range(100, 50)         select new {Number = n, OddEven = n % 2 == 1 ? "odd" : "even"}; The above query will generate 50 records where the record will start from 100 till 149. The query also determines if the number is odd or even. But if we want to generate the same number for multiple times then we can use the Repeat method.     var numbers = Enumerable.Repeat(7, 10); The above query will produce a list with 10 items having the value 7. Vikram

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  • Operator of the Week - Spools, Eager Spool

    For the fifth part of Fabiano's mission to describe the major Showplan Operators used by SQL Server's Query Optimiser, he introduces the spool operators and particularly the Eager Spool, explains blocking and non-blocking and then describes how the Halloween Problem is avoided.

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  • Showplan Operator of the week - Assert

    As part of his mission to explain the Query Optimiser in practical terms, Fabiano attempts the feat of describing, one week at a time, all the major Showplan Operators used by SQL Server's Query Optimiser to build the Query Plan. He starts with Assert

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  • Ruby: if statement using regexp and boolean operator [migrated]

    - by bev
    I'm learning Ruby and have failed to make a compound 'if' statement work. Here's my code (hopefully self explanatory) commentline = Regexp.new('^;;') blankline = Regexp.new('^(\s*)$') if (line !~ commentline || line !~ blankline) puts line end the variable 'line' is gotten from reading the following file: ;; alias filename backupDir Prog_i Prog_i.rb ./store Prog_ii Prog_ii.rb ./store This fails and I'm not sure why. Basically I want the comment lines and blank lines to be ignored during the processing of the lines in the file. Thanks for your help.

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  • Showplan Operator of the Week – BookMark/Key Lookup

    Fabiano continues in his mission to describe the major Showplan Operators used by SQL Server's Query Optimiser. This week he meets a star, the Key Lookup, a stalwart performer, but most famous for its role in ill-performing queries where an index does not 'cover' the data required to execute the query. If you understand why, and in what circumstances, key lookups are slow, it helps greatly with optimising query performance.

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  • How do I overload an operator for an enumeration in C#?

    - by ChrisHDog
    I have an enumerated type that I would like to define the , <, =, and <= operators for. I know that these operators are implictly created on the basis of the enumerated type (as per the documentation) but I would like to explictly define these operators (for clarity, for control, to know how to do it, etc...) I was hoping I could do something like: public enum SizeType { Small = 0, Medium = 1, Large = 2, ExtraLarge = 3 } public SizeType operator >(SizeType x, SizeType y) { } But this doesn't seem to work ("unexpected toke") ... is this possible? It seems like it should be since there are implictly defined operators. Any suggestions?

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  • How does delete deal with pointer constness?

    - by aJ
    I was reading this question Deleting a const pointer and wanted to know more about delete behavior. Now, as per my understanding: delete expression works in two steps: invoke destructor then releases the memory (often with a call to free()) by calling operator delete. operator delete accepts a void*. As part of a test program I overloaded operator delete and found that operator delete doesn't accept const pointer. Since operator delete does not accept const pointer and delete internally calls operator delete, how does Deleting a const pointer work ? Does delete uses const_cast internally?

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  • Why would the assignment operator ever do something different than its matching constructor?

    - by Neil G
    I was reading some boost code, and came across this: inline sparse_vector &assign_temporary(sparse_vector &v) { swap(v); return *this; } template<class AE> inline sparse_vector &operator=(const sparse_vector<AE> &ae) { self_type temporary(ae); return assign_temporary(temporary); } It seems to be mapping all of the constructors to assignment operators. Great. But why did C++ ever opt to make them do different things? All I can think of is scoped_ptr?

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  • Why does javascript's "in" operator return true when testing if 0 exists in an array that doesn't co

    - by Mariano Peterson
    For example, this returns true, and makes sense: var x = [1,2]; 1 in x; // true This returns false, and makes sense: var x = [1,2]; 3 in x; // false However this returns true, and I don't understand why: var x = [1,2]; 0 in x; You can quickly test it by putting this in your browser's address bar: javascript:var x=[1,2]; alert(0 in x); Why does the "in" operator in Javascript return true when testing if "0" exists in array, even when the array doesn't appear to contain "0"?

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  • C++: Overload != When == Overloaded

    - by Mark W
    Say I have a class where I overloaded the operator == as such: Class A { ... public: bool operator== (const A &rhs) const; ... }; ... bool A::operator== (const A &rhs) const { .. return isEqual; } I already have the operator == return the proper Boolean value. Now I want to extend this to the simple opposite (!=). I would like to call the overloaded == operator and return the opposite, i.e. something of the nature bool A::operator!= (const A &rhs) const { return !( this == A ); } Is this possible? I know this will not work, but it exemplifies what I would like to have. I would like to keep only one parameter for the call: rhs. Any help would be appreciated, because I could not come up with an answer after several search attempts.

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  • How do I overload the square-bracket operator in C#?

    - by Coderer
    DataGridView, for example, lets you do this: DataGridView dgv = ...; DataGridViewCell cell = dgv[1,5]; but for the life of me I can't find the documentation on the index/square-bracket operator. What do they call it? Where is it implemented? Can it throw? How can I do the same thing in my own classes? ETA: Thanks for all the quick answers. Briefly: the relevant documentation is under the "Item" property; the way to overload is by declaring a property like public object this[int x, int y]{ get{...}; set{...} }; the indexer for DataGridView does not throw, at least according to the documentation. It doesn't mention what happens if you supply invalid coordinates. ETA Again: OK, even though the documentation makes no mention of it (naughty Microsoft!), it turns out that the indexer for DataGridView will in fact throw an ArgumentOutOfRangeException if you supply it with invalid coordinates. Fair warning.

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  • How do I overload () operator with two parameters; like (3,5)?

    - by hkBattousai
    I have a mathematical matrix class. It contains a member function which is used to access any element of the class. template >class T> class Matrix { public: // ... void SetElement(T dbElement, uint64_t unRow, uint64_t unCol); // ... }; template <class T> void Matrix<T>::SetElement(T Element, uint64_t unRow, uint64_t unCol) { try { // "TheMatrix" is define as "std::vector<T> TheMatrix" TheMatrix.at(m_unColSize * unRow + unCol) = Element; } catch(std::out_of_range & e) { // Do error handling here } } I'm using this method in my code like this: // create a matrix with 2 rows and 3 columns whose elements are double Matrix<double> matrix(2, 3); // change the value of the element at 1st row and 2nd column to 6.78 matrix.SetElement(6.78, 1, 2); This works well, but I want to use operator overloading to simplify things, like below: Matrix<double> matrix(2, 3); matrix(1, 2) = 6.78; // HOW DO I DO THIS?

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  • Operator precedence and struct definition in C

    - by Yktula
    struct struct0 { int a; }; struct struct1 { struct struct0 structure0; int b; } rho; &rho->structure0; /* Reference 1 */ (struct struct0 *)rho; /* Reference 2 */ (struct struct0)rho; /* Reference 3 */ From reference 1, does the compiler take the address of rho, and then access structure0, or vice-versa? What does the line at reference 2 do? Since structure0 is the first member of struct1, would reference 3 be equivalent to reference 1?

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  • On ocamlyacc, function application grammar and precedence

    - by Amadan
    I'm OCaml newbie and I'm trying to write a simple OCaml-like grammar, and I can't figure this out. My grammar allows something like this: let sub = fun x -> fun y -> x - y;; However, if I want to use the function so defined, I can write: (sub 7) 3 but I can't write sub 7 3, which really bugs me. For some reason, it gets interpreted as if I wrote sub (7 3) (which would treat 7 as a function with argument 3). The relevant sections are: /* other operators, then at the very end: */ %left APPLY /* ... */ expr: /* ... */ | expr expr %prec APPLY { Apply($1, $2) } Thanks!

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  • cons operator (::) in F#

    - by Max
    The :: operator in F# always prepends elements to the list. Is there an operator that appends to the list? I'm guessing that using @ operator [1; 2; 3] @ [4] would be less efficient, than appending one element.

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  • How to make that the LanguageBinder take precedence over the DynamicBinder

    - by rudimenter
    Hi I Have a class which implement IDynamicMetaObjectProvider I implement the BindGetMember Method from DynamicMetaObject. Now when i Generate a dynamic Object and Access a property every call gets implicit passed through the BindGetMember Method. I want that at first the language Binder get his chance before my code comes in. It is somehow doable with "binder.FallbackGetMember" but i am not sure how the expression has to look like. I call here dynamic com=CommandFactory.GetCommand(); com.testprop; //expected: "test"; but "test2" comes back public class Command : System.Dynamic.IDynamicMetaObjectProvider { public string testprop { get { return "test"; } } public object GetValue(string name) { return "test2"; } System.Dynamic.DynamicMetaObject System.Dynamic.IDynamicMetaObjectProvider.GetMetaObject(System.Linq.Expressions.Expression parameter) { return new MetaCommand(parameter, this); } private class MetaCommand : System.Dynamic.DynamicMetaObject { public MetaCommand(Expression expression, Command value) : base(expression, System.Dynamic.BindingRestrictions.Empty, value) { } public override System.Dynamic.DynamicMetaObject BindGetMember(System.Dynamic.GetMemberBinder binder) { var self = this.Expression; var bag = (Command)base.Value; Expression target; target = Expression.Call( Expression.Convert(self, typeof(Command)), typeof(Command).GetMethod("GetValue"), Expression.Constant(binder.Name) ); var restrictions = BindingRestrictions .GetInstanceRestriction(self, bag); return new DynamicMetaObject(target, restrictions); } #endregion } }

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