Search Results

Search found 2508 results on 101 pages for 'ternary operator'.

Page 13/101 | < Previous Page | 9 10 11 12 13 14 15 16 17 18 19 20  | Next Page >

  • 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

    Read the article

  • 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?

    Read the article

  • 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();          } }

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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"?

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • How to use ternary operator instead of if-else in PHP

    - by Mac Taylor
    hey guys i need to shorten or better to say ., harden my codes this is my original code : if ($type = "recent") { $OrderType = "sid DESC"; }elseif ($type = "pop"){ $OrderType = "counter DESC"; }else { $OrderType = "RAND()"; } now how can i use markers like this : $OrderType = ($type = "recent") ? "sid DESC" : "counter DESC" ; i tried but i didnt know how to write elseif in operators

    Read the article

  • Logical operator AND having higher order of precedence than IN

    - by AspOnMyNet
    I’ve read that logical operator AND has higher order of precedence than logical operator IN, but that doesn’t make sense since if that was true, then wouldn’t in the following statement the AND condition got evaluated before the IN condition ( thus before IN operator would be able to check whether Released field equals to any of the values specified within parentheses ? SELECT Song, Released, Rating FROM Songs WHERE Released IN (1967, 1977, 1987) AND SongName = ’WTTJ’ thanx

    Read the article

  • HQL over ternary map with subcollection

    - by Diego Mijelshon
    I'm stuck with a query I need to write. Given the following model: public class A : Entity<Guid> { public virtual IDictionary<B, C> C { get; set; } } public class B : Entity<Guid> { } public class C : Entity<Guid> { public virtual int Data1 { get; set; } public virtual ICollection<D> D { get; set; } } public class D : Entity<Guid> { public virtual int Data2 { get; set; } } I need to get a list of A instances that have a D containing some data for the specified B (parameter) In the object model, that would be: listOfA.Where(a => a.C[b].D.Any(d => d.Data2 == 0)) But I wasn't able to write a working HQL. I'm able to write something like the following, which filters on C.Data1: from A a where a.C[:b].Data1 = 0 But I'm unable to do anything with the elements of a.C[:b].D (I get various parsing exceptions) Here are the mappings, in case you're interested (nothing special, generated by ConfORM): <class name="A"> <id name="Id" type="Guid"> <generator class="guid.comb" /> </id> <map name="C"> <key column="a_key" /> <map-key-many-to-many class="B" /> <one-to-many class="C" /> </map> </class> <class name="B"> <id name="Id" type="Guid"> <generator class="guid.comb" /> </id> </class> <class name="C"> <id name="Id" type="Guid"> <generator class="guid.comb" /> </id> <property name="Data1" /> <bag name="D"> <key column="c_key" /> <one-to-many class="D" /> </bag> </class> <class name="D"> <id name="Id" type="Guid"> <generator class="guid.comb" /> </id> <property name="Data2" /> </class> Thanks!

    Read the article

  • Error in ternary expression

    - by Bipul
    Consider the following code which shows compile time error : #include <stdio.h> int main(int argc, char** argv) { int x=5,y=0,z=2; int a=z?x,y?x:(y); // but z?x,y?x:y:z is not showing any error printf("%d",a); return 0; } Please help me explain the reason why z?x,y?x:y:z is not showing any error?

    Read the article

  • 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.

    Read the article

< Previous Page | 9 10 11 12 13 14 15 16 17 18 19 20  | Next Page >