Search Results

Search found 5165 results on 207 pages for 'const cast'.

Page 2/207 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • C function const multidimensional-array argument strange warning

    - by rogi
    Ehllo, I'm getting some strange warning about this code: typedef double mat4[4][4]; void mprod4(mat4 r, const mat4 a, const mat4 b) { /* yes, function is empty */ } int main() { mat4 mr, ma, mb; mprod4(mr, ma, mb); } gcc output as follows: $ gcc -o test test.c test.c: In function 'main': test.c:13: warning: passing argument 2 of 'mprod4' from incompatible pointer type test.c:4: note: expected 'const double (*)[4]' but argument is of type 'double (*)[4]' test.c:13: warning: passing argument 3 of 'mprod4' from incompatible pointer type test.c:4: note: expected 'const double ()[4]' but argument is of type 'double ()[4]' defining the function as: void mprod4(mat4 r, mat4 a, mat4 b) { } OR defining matrices at main as: mat4 mr; const mat4 ma; const mat4 mb; OR calling teh function in main as: mprod4(mr, (const double(*)[4])ma, (const double(*)[4])mb); OR even defining mat4 as: typedef double mat4[16]; make teh warning go away. Wat is happening here? Am I doing something invalid? gcc version is 4.4.3 if relevant. Thanks for your attention.

    Read the article

  • Detecting const-ness of nested type

    - by Channel72
    Normally, if I need to detect whether a type is const I just use boost::is_const. However, I ran into trouble when trying to detect the const-ness of a nested type. Consider the following traits template, which is specialized for const types: template <class T> struct traits { typedef T& reference; }; template <class T> struct traits<const T> { typedef T const& reference; }; The problem is that boost::is_const doesn't seem to detect that traits<const T>::reference is a const type. For example: std::cout << std::boolalpha; std::cout << boost::is_const<traits<int>::reference>::value << " "; std::cout << boost::is_const<traits<const int>::reference>::value << std::endl; This outputs: false false Why doesn't it output false true?

    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

  • Returning a C++ reference in a const member functionasses

    - by Chris Kaminski
    A have a class hierarchy that looks somethign like this: class AbstractDataType { public: virtual int getInfo() = 0; }; class DataType: public AbstractDataType { public: virtual int getInfo() { }; } class Accessor { DataType data; public: const AbstractDataType& getData() const { return(data); } } Well, GCC 4.4 reports: In member function ‘const AbstractDataType& Accessor::getData() const’: error: invalid initialization of reference of type ‘const AbstractDataType&’ from expression of type ‘const DataType’ Where am I going wrong - is this a case where I MUST use a pointer?

    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

  • C++: Why does gcc prefer non-const over const when accessing operator[]?

    - by JonasW
    This question might be more appropriately asked regarding C++ in general, but as I am using gcc on linux that's the context. Consider the following program: #include <iostream> #include <map> #include <string> using namespace std; template <typename TKey, typename TValue> class Dictionary{ public: map<TKey, TValue> internal; TValue & operator[](TKey const & key) { cout << "operator[] with key " << key << " called " << endl; return internal[key]; } TValue const & operator[](TKey const & key) const { cout << "operator[] const with key " << key << " called " << endl; return internal.at(key); } }; int main(int argc, char* argv[]) { Dictionary<string, string> dict; dict["1"] = "one"; cout << "first one: " << dict["1"] << endl; return 0; } When executing the program, the output is: operator[] with key 1 called operator[] with key 1 called first one: one What I would like is to have the compiler choose the operator[]const method instead in the second call. The reason is that without having used dict["1"] before, the call to operator[] causes the internal map to create the data that does not exist, even if the only thing I wanted was to do some debugging output, which of course is a fatal application error. The behaviour I am looking for would be something like the C# index operator which has a get and a set operation and where you could throw an exception if the getter tries to access something that doesn't exist: class MyDictionary<TKey, TVal> { private Dictionary<TKey, TVal> dict = new Dictionary<TKey, TVal>(); public TVal this[TKey idx] { get { if(!dict.ContainsKey(idx)) throw KeyNotFoundException("..."); return dict[idx]; } set { dict[idx] = value; } } } Thus, I wonder why the gcc prefers the non-const call over the const call when non-const access is not required.

    Read the article

  • Cast vector<T> to vector<const T>

    - by user345386
    I have a member variable of type vector (where is T is a custom class, but it could be int as well.) I have a function from which I want to return a pointer to this vector, but I don't want the caller to be able to change the vector or it's items. So I want the return type to be const vector* None of the casting methods I tried worked. The compiler keeps complaining that T is not compatible with const T. Here's some code that demonstrates the gist of what I'm trying to do; vector<int> a; const vector<const int>* b = (const vector<const int>* ) (&a); This code doesn't compile for me. Thanks in advance!

    Read the article

  • Is it ok to dynamic cast "this" as a return value?

    - by Panayiotis Karabassis
    This is more of a design question. I have a template class, and I want to add extra methods to it depending on the template type. To practice the DRY principle, I have come up with this pattern (definitions intentionally omitted): template <class T> class BaseVector: public boost::array<T, 3> { protected: BaseVector<T>(const T x, const T y, const T z); public: bool operator == (const Vector<T> &other) const; Vector<T> operator + (const Vector<T> &other) const; Vector<T> operator - (const Vector<T> &other) const; Vector<T> &operator += (const Vector<T> &other) { (*this)[0] += other[0]; (*this)[1] += other[1]; (*this)[2] += other[2]; return *dynamic_cast<Vector<T> * const>(this); } } template <class T> class Vector : public BaseVector<T> { public: Vector<T>(const T x, const T y, const T z) : BaseVector<T>(x, y, z) { } }; template <> class Vector<double> : public BaseVector<double> { public: Vector<double>(const double x, const double y, const double z); Vector<double>(const Vector<int> &other); double norm() const; }; I intend BaseVector to be nothing more than an implementation detail. This works, but I am concerned about operator+=. My question is: is the dynamic cast of the this pointer a code smell? Is there a better way to achieve what I am trying to do (avoid code duplication, and unnecessary casts in the user code)? Or am I safe since, the BaseVector constructor is private?

    Read the article

  • const object and const constructor

    - by Muhammad alaa
    Is there any way to know if an object is a const object or regular object, for instance consider the following class class String { String(const char* str); }; if user create a const object from String then there is no reason to copy the passed native string and that because he will not make any manipulation on it, the only thing he will do is get string size, string search and other functions that will not change the string.

    Read the article

  • invalid conversion from ‘float**’ to ‘const float**’

    - by Omry
    I have a function that receives float** as an argument, and I tried to change it to take const float**. the compiler (g++) didn't like it and issued : invalid conversion from ‘float**’ to ‘const float**’ this makes no sense to me, I know (and verified) that I can pass char* to a function that takes const char*, so why not with const float** ?

    Read the article

  • c++ meaning of the use of const in the signature

    - by jbu
    Please help me understand the following signature: err_type funcName(const Type& buffer) const; so for the first const, does that mean the contents of Type cannot change or that the reference cannot change? secondly, what does the second const mean? I don't really even have a hint. Thanks in advance, jbu

    Read the article

  • is const (c++) optional?

    - by Dr Deo
    according to some tutorials i read a while back, the "const" declaration makes a variable "constant" ie it cannot change later. But i find this const declaration abit inconveniencing since the compiler sometimes gives errors like "cannot convert const int to int" or something like that. and i find myself cheating by removing it anyway. question: assuming that i am careful about not changing a variable in my source code, can i happily forget about this const stuff? Thanks in advance

    Read the article

  • RAII: Initializing data member in const method

    - by Thomas Matthews
    In RAII, resources are not initialized until they are accessed. However, many access methods are declared constant. I need to call a mutable (non-const) function to initialize a data member. Example: Loading from a data base struct MyClass { int get_value(void) const; private: void load_from_database(void); // Loads the data member from database. int m_value; }; int MyClass :: get_value(void) const { static bool value_initialized(false); if (!value_initialized) { // The compiler complains about this call because // the method is non-const and called from a const // method. load_from_database(); } return m_value; } My primitive solution is to declare the data member as mutable. I would rather not do this, because it suggests that other methods can change the member. How would I cast the load_from_database() statement to get rid of the compiler errors?

    Read the article

  • Using a class with const data members in a vector

    - by Max
    Given a class like this: class Foo { const int a; }; Is it possible to put that class in a vector? When I try, my compiler tells me it can't use the default assignment operator. I try to write my own, but googling around tells me that it's impossible to write an assignment operator for a class with const data members. One post I found said that "if you made [the data member] const that means you don't want assignment to happen in the first place." This makes sense. I've written a class with const data members, and I never intended on using assignment on it, but apparently I need assignment to put it in a vector. Is there a way around this that still preserves const-correctness?

    Read the article

  • Use of const double for intermediate results

    - by Arne
    Hi, I a writing a Simulation program and wondering if the use of const double is of any use when storing intermediate results. Consider this snippet: double DoSomeCalculation(const AcModel &model) { (...) const double V = model.GetVelocity(); const double m = model.GetMass(); const double cos_gamma = cos(model.GetFlightPathAngleRad()); (...) return m*V*cos_gamma*Chi_dot; } Note that the sample is there only to illustrate -- it might not make to much sense from the engineering side of things. The motivation of storing for example cos_gamma in a variable is that this cosine is used many time in other expressions covered by (...) and I feel that the code gets more readable when using cos_gamma rather than cos(model.GetFlightPathAngleRad()) in various expressions. Now the actual is question is this: since I expect the cosine to be the same througout the code section and I actually created the thing only as a placeholder and for convenience I tend to declare it const. Is there a etablished opinion on wether this is good or bad practive or whether it might bite me in the end? Does a compiler make any use of this additional information or am I actually hindering the compiler from performing useful optimizations? Arne

    Read the article

  • Const operator overloading problems in C++

    - by steigers
    Hello everybody, I'm having trouble with overloading operator() with a const version: #include <iostream> #include <vector> using namespace std; class Matrix { public: Matrix(int m, int n) { vector<double> tmp(m, 0.0); data.resize(n, tmp); } ~Matrix() { } const double & operator()(int ii, int jj) const { cout << " - const-version was called - "; return data[ii][jj]; } double & operator()(int ii, int jj) { cout << " - NONconst-version was called - "; if (ii!=1) { throw "Error: you may only alter the first row of the matrix."; } return data[ii][jj]; } protected: vector< vector<double> > data; }; int main() { try { Matrix A(10,10); A(1,1) = 8.8; cout << "A(1,1)=" << A(1,1) << endl; cout << "A(2,2)=" << A(2,2) << endl; double tmp = A(3,3); } catch (const char* c) { cout << c << endl; } } This gives me the following output: NONconst-version was called - - NONconst-version was called - A(1,1)=8.8 NONconst-version was called - Error: you may only alter the first row of the matrix. How can I achieve that C++ call the const-version of operator()? I am using GCC 4.4.0. Thanks for your help! Sebastian

    Read the article

  • const correctness

    - by ra170
    I was going through: C++ FAQs about inheritance and decided to implement it (just to learn it) #include "Shape.h" void Shape::print() const { float a = this->area(); // area() is pure virtual ... } now, everything (well, almost) works as described in item: faq:23.1 except that print() is const and so it can't access the "this" pointer, as soon as you take out const, it works. Now, C++ FAQs have been around for a while and are usually pretty good. Is this a mistake? Do they have typo or am I wrong? If I'm wrong, I would like to know how is it possible to access the "this" pointer in a const function.

    Read the article

  • An operator == whose parameters are non-const references

    - by Eduardo León
    I this post, I've seen this: class MonitorObjectString: public MonitorObject { // some other declarations friend inline bool operator==(/*const*/ MonitorObjectString& lhs, /*const*/ MonitorObjectString& rhs) { return lhs.fVal==rhs.fVal; } } Before we can continue, THIS IS VERY IMPORTANT: I am not questioning anyone's ability to code. I am just wondering why someone would need non-const references in a comparison. The poster of that question did not write that code. This was just in case. This is important too: I added both /*const*/s and reformatted the code. Now, we get back to the topic: I can't think of a sane use of the equality operator that lets you modify its by-ref arguments. Do you?

    Read the article

  • What effect does static const have on a namespace member

    - by user144182
    namespace MyNamespace { static const double GasConstant = 1.987; Class MyClass { // constructors, methods, etc. }; }; I previously had GasConstant declared within the MyClass declaration (and had a separate definition in the source file since C++ does not support const initialization of non-integral types). I however need to access it from other files and also logically it seems like it should reside at the namespace level. My questions is, what effect does static const have in this case? Clearly const means I can't assign a new value to GasConstant, but what does a static member at the namespace mean. Is this similar to filescope static effect, where the member is not accessible outside of the unit?

    Read the article

  • Const parameter at constructor causes stackoverflow

    - by Luca
    I've found this strange behavior with VS2005 C++ compiler. Here is the situation: I cannot publish the code, but situation is very simple. Here is initial code: it work perfectly class Foo { public: Foo(Bar &bar) { ... } } The constructor implementation stores a reference, setup some members... indeed nothing special. If I change the code in the following way: class Foo { public: Foo(const Bar &bar) { ... } } I've added a const qualifier to the only constructor routine parameter. It compiles correctly, but the compiler outputs a warning saying that the routine Foo::Foo will cause a stackoverflow (even if the execution path doesn't construct any object Foo); effectively this happens. So, why the code without the const parameter works perfectly, while the one with the const qualifier causes a stackoverflow? What can cause this strange behavior?

    Read the article

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