Search Results

Search found 977 results on 40 pages for 'operators'.

Page 12/40 | < Previous Page | 8 9 10 11 12 13 14 15 16 17 18 19  | Next Page >

  • What does the symbol :=: mean

    - by Dan Maguire
    I've found the symbol :=: in some Clarion code and I can't seem to figure out exactly what it does. The code was written by a previous developer many years ago, so I can't ask him. I also have not been able to find any results for "colon equals colon" in Google. Here is an example of the code, where bufSlcdpaDtl is a file object: lCCRecord Like(bufSlcdpaDtl),Pre(lCCRecord) ! ...other stuff... lCCRecord :=: bufSlcdpaDtl I'm wondering if it's something similar to ::= in Python or possibly the assignment operator :=.

    Read the article

  • Swig C++ Lua Pass class by reference

    - by Jeremy
    I don't know why I'm having a hard time with this. All I want to do is this: class foo { public: foo(){} ~foo(){} float a,b; }; class foo2 { public: foo2(){} foo2(const foo &f){*this = f;} ~foo2(){} void operator=(const foo& f){ x = f.a; y = f.b; } float x,y; }; /* Usage(cpp): foo f; foo2 f2(f); //or using the = operator f2 = f; */ The problem I'm having is that, after swigging this code, I can't figure out how to make the lua script play nice. /* Usage(lua) f = example.foo() f2 = example.foo2(f) --error */ The error I get is "Wrong arguments for overloaded function 'new_Foo2'": Possible c/c++ prototypes are: foo2() foo2(foo const &) The same thing happens if I try and use do f2 = f. As I understand it everything is stored as a pointer so I did try adding an additional constructor that took a pointer to foo but to no avail.

    Read the article

  • 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).

    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

  • printing using one '\n'

    - by Alex
    I am pretty sure all of you are familiar with the concept of the Big4, and I have several stuffs to do print in each of the constructor, assignment, destructor, and copy constructor. The restriction is this: I CAN'T use more than one newline (e.g., ƒn or std::endl) in any method I can have a method called print, so I am guessing print is where I will put that precious one and only '\n', my problem is that how can the method print which prints different things on each of the element I want to print in each of the Big4? Any idea? Maybe overloading the Big4?

    Read the article

  • C++ overide global operator comma gives error

    - by uray
    the second function gives error C2803 http://msdn.microsoft.com/en-us/library/zy7kx46x%28VS.80%29.aspx : 'operator ,' must have at least one formal parameter of class type. any clue? template<class T,class A = std::allocator<T>> class Sequence : public std::vector<T,A> { public: Sequence<T,A>& operator,(const T& a) { this->push_back(a); return *this; } Sequence<T,A>& operator,(const Sequence<T,A>& a) { for(Sequence<T,A>::size_type i=0 ; i<a.size() ; i++) { this->push_back(a.at(i)); } return *this; } }; //this works! template<typename T> Sequence<T> operator,(const T& a, const T&b) { Sequence<T> seq; seq.push_back(a); seq.push_back(b); return seq; } //this gives error C2803! Sequence<double> operator,(const double& a, const double& b) { Sequence<double> seq; seq.push_back(a); seq.push_back(b); return seq; }

    Read the article

  • C++ overloading operator comma for variadic arguments

    - by uray
    is it possible to construct variadic arguments for function by overloading operator comma of the argument? i want to see an example how to do so.., maybe something like this: template <typename T> class ArgList { public: ArgList(const T& a); ArgList<T>& operator,(const T& a,const T& b); } //declaration void myFunction(ArgList<int> list); //in use: myFunction(1,2,3,4); //or maybe: myFunction(ArgList<int>(1),2,3,4);

    Read the article

  • C# XOR on two byte variables will not compile without a cast

    - by Ash
    Why does the following raise a compile time error: 'Cannot implicitly convert type 'int' to 'byte': byte a = 25; byte b = 60; byte c = a ^ b; This would make sense if I were using an arithmentic operator because the result of a + b could be larger than can be stored in a single byte. However applying this to the XOR operator is pointless. XOR here it a bitwise operation that can never overflow a byte. using a cast around both operands works: byte c = (byte)(a ^ b);

    Read the article

  • Performance difference in for loop condition?

    - by CSharperWithJava
    Hello all, I have a simple question that I am posing mostly for my curiousity. What are the differences between these two lines of code? (in C++) for(int i = 0; i < N, N > 0; i++) for(int i = 0; i < N && N > 0; i++) The selection of the conditions is completely arbitrary, I'm just interested in the differences between , and &&. I'm not a beginner to coding by any means, but I've never bothered with the comma operator. Are there performance/behavior differences or is it purely aesthetic? One last note, I know there are bigger performance fish to fry than a conditional operator, but I'm just curious. Indulge me. Edit Thanks for your answers. It turns out the code that prompted this question had misused the comma operator in the way I've described. I wondered what the difference was and why it wasn't a && operator, but it was just written incorrectly. I didn't think anything was wrong with it because it worked just fine. Thanks for straightening me out.

    Read the article

  • what does this C++ line of code mean "sol<?=f((1<<n)-1,i,0)+abs(P[i])*price;"

    - by KItis
    Could anyone help me to understand following line of code. sol I am studying an algorithm written using c++ and it has following operator " following is the error message returned. Hello.cpp: In function ‘int main()’: Hello.cpp:115: error: ‘memset’ was not declared in this scope Hello.cpp:142: error: expected primary-expression before ‘?’ token Hello.cpp:142: error: expected primary-expression before ‘=’ token Hello.cpp:142: error: expected ‘:’ before ‘;’ token Hello.cpp:142: error: expected primary-expression before ‘;’ token may be " Thanks in advance for the time you spent reading this post.

    Read the article

  • Please explain this php expression "!$variable"

    - by DogBot
    What does an exclamaton mark in front of a variable mean? And how is it being used in this piece of code? EDIT: From the answers so far I suspect that I also should mention that this code is in a function where one of the parameters is $mytype ....would this be a way of checking if $mytype was passed? - Thanks to all of the responders so far. $myclass = null; if ($mytype == null && ($PAGE->pagetype <> 'site-index' && $PAGE->pagetype <>'admin-index')) { return $myclass; } elseif ($mytype == null && ($PAGE->pagetype == 'site-index' || $PAGE->pagetype =='admin-index')) { $myclass = ' active_tree_node'; return $myclass; } elseif (!$mytype == null && ($PAGE->pagetype == 'site-index' || $PAGE->pagetype =='admin-index')) { return $myclass; }`

    Read the article

  • Implementing operator< in C++

    - by Vulcan Eager
    I have a class with a few numeric fields such as: class Class1 { int a; int b; int c; public: // constructor and so on... bool operator<(const Class1& other) const; }; I need to use objects of this class as a key in an std::map. I therefore implement operator<. What is the simplest implementation of operator< to use here?

    Read the article

  • Algorithm for bitwise fiddling

    - by EquinoX
    If I have a 32-bit binary number and I want to replace the lower 16-bit of the binary number with a 16-bit number that I have and keep the upper 16-bit of that number to produce a new binary number.. how can I do this using simple bitwise operator? For example the 32-bit binary number is: 1010 0000 1011 1111 0100 1000 1010 1001 and the lower 16-bit I have is: 0000 0000 0000 0001 so the result is: 1010 0000 1011 1111 0000 0000 0000 0001 how can I do this?

    Read the article

  • How to make += operator keep the object reference?

    - by orloffm
    Say, I have a class: class M { public int val; And also a + operator inside it: public static M operator +(M a, M b) { M c = new M(); c.val = a.val + b.val; return c; } } And I've got a List of the objects of the class: List<M> ms = new List(); M obj = new M(); obj.val = 5; ms.Add(obj); Some other object: M addie = new M(); addie.val = 3; I can do this: ms[0] += addie; and it surely works as I expect - the value in the list is changed. But if I want to do M fromList = ms[0]; fromList += addie; it doesn't change the value in ms for obvious reasons. But intuitively I expect ms[0] to also change after that. Really, I pick the object from the list and then I increase it's value with some other object. So, since I held a reference to ms[0] in fromList before addition, I want still to hold it in fromList after performing it. Are there any ways to achieve that?

    Read the article

  • How can I access the sign bit of a number in C++?

    - by Keand64
    I want to be able to access the sign bit of a number in C++. My current code looks something like this: int sign bit = number >> 31; That appears to work, giving me 0 for positive numbers and -1 for negative numbers. However, I don't see how I get -1 for negative numbers: if 12 is 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 then -12 is 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 and shifting it 31 bits would make 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 which is 1, not -1, so why do I get -1 when I shift it?

    Read the article

  • C# implicit conversions and == operator

    - by Arnis L.
    Some code for context: class a { } class b { public a a{get;set;} public static implicit operator a(b b) { return b.a; } } a a=null; b b=null; a = b; //compiler: cannot apply operator '==' to operands of type tralala... bool c = a == b; Is it possible to use == operator on different type instances, where one can implicitly convert to another? What did i miss? Edit: If types must be the same calling ==, then why int a=1; double b=1; bool c=a==b; works?

    Read the article

  • How to change the meaning of pointer access operator

    - by kumar_m_kiran
    Hi All, This may be very obvious question, pardon me if so. I have below code snippet out of my project, #include <stdio.h> class X { public: int i; X() : i(0) {}; }; int main(int argc,char *arv[]) { X *ptr = new X[10]; unsigned index = 5; cout<<ptr[index].i<<endl; return 0; } Question Can I change the meaning of the ptr[index] ? Because I need to return the value of ptr[a[index]] where a is an array for subindexing. I do not want to modify existing source code. Any new function added which can change the behavior is needed. Since the access to index operator is in too many places (536 to be precise) in my code, and has complex formulas inside the index subscript operator, I am not inclined to change the code in many locations. PS : 1. I tried operator overload and came to conclusion that it is not possible. 2. Also p[i] will be transformed into *(p+i). I cannot redefine the basic operator '+'. So just want to reconfirm my understanding and if there are any possible short-cuts to achieve. Else I need fix it by royal method of changing every line of code :) .

    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

  • undefined C/C++ symbol as operator

    - by uray
    I notice that the character/symbol '`' and '@' is not used as an operator in C/C++, does anyone know the reason or historically why its so? if its really not used, is it safe to define those symbols as another operator/statement using #define?

    Read the article

  • Is there an exponent operator in C#?

    - by Charlie
    For example, does an operator exist to handle this? float Result, Number1, Number2; Number1 = 2; Number2 = 2; Result = Number1 (operator) Number2; In the past the ^ operator has served as an exponential operator in other languages, but in C# it is a bit-wise operator. Do I have to write a loop or include another namespace to handle exponential operations? If so, how do I handle exponential operations using non-integers?

    Read the article

  • difference between -> and . for member selection operator.

    - by TimothyTech
    in this book i have I'm learning pointers, and i just got done with the chapter about OOP (spits on ground) anyways its telling me i can use a member selection operator like this ( - ). it sayd that is is like the "." except points to objects rather than member objects. whats the difference, it looks like it is used the same way...

    Read the article

< Previous Page | 8 9 10 11 12 13 14 15 16 17 18 19  | Next Page >