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  • 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.

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  • 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?

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  • 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?

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  • 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.

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  • 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!

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  • Const Functions and Interfaces in C++

    - by 58gh1z
    I'll use the following (trivial) interface as an example: struct IObject { virtual ~IObject() {} virtual std::string GetName() const = 0; virtual void ChangeState() = 0; }; Logic dictates that GetName should be a const member function while ChangeState shouldn't. All code that I've seen so far doesn't follow this logic, though. That is, GetName in the example above wouldn't be marked as a const member function. Is this laziness/carelessness or is there a legitimate reason for this? What are the major cons of me forcing my clients to implement const member functions when they are logically called for?

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  • 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.

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  • 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

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  • 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

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  • 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?

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  • 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

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  • 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

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  • 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?

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  • 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?

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  • 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?

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  • Cannot initialize non-const reference from convertible type

    - by Julien L.
    Hi, I cannot initialize a non-const reference to type T1 from a convertible type T2. However, I can with a const reference. long l; const long long &const_ref = l; // fine long long &ref = l; // error: invalid initialization of reference of // type 'long long int&' from expression of type // 'long int' Most problems I encountered were related to r-values that cannot be assigned to a non-const reference. This is not the case here -- can someone explain? Thanks.

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  • How to const declare the this pointer sent as parameter

    - by Tomas
    Hi, I want to const declare the this pointer received as an argument. static void Class::func(const OtherClass *otherClass) { // use otherClass pointer to read, but not write to it. } It is being called like this: void OtherClass::func() { Class::func(this); } This does not compile nad if i dont const declare the OtherClass pointer, I can change it. Thanks.

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  • How to iterate properly across a const set?

    - by Jared
    I'm working on a program that's supposed to represent a graph. My issue is in my printAdjacencyList function. Basically, I have a Graph ADT that has a member variable "nodes", which is a map of the nodes of that graph. Each Node has a set of Edge* to the edges it is connected to. I'm trying to iterate across each node in the graph and each edge of a node. void MyGraph::printAdjacencyList() { std::map<std::string, MyNode*>::iterator mit; std::set<MyEdge*>::iterator sit; for (mit = nodes.begin(); mit != nodes.end(); mit++ ) { std::cout << mit->first << ": {"; const std::set<MyEdge*> edges = mit->second->getEdges(); for (sit = edges.begin(); sit != edges.end(); sit++) { std::pair<MyNode*, MyNode*> edgeNodes = *sit->getEndpoints(); } } std::cout << " }" << std::endl; } getEdges is declared as: const std::set<MyEdge*>& getEdges() { return edges; }; and get Endpoints is declared as: const std::pair<MyNode*, MyNode*>& getEndpoints() { return nodes; }; The compiler error I'm getting is: MyGraph.cpp:63: error: request for member `getEndpoints' in `*(&sit)->std::_Rb_tree_const_iterator<_Tp>::operator-> [with _Tp = MyEdge*]()', which is of non-class type `MyEdge* const' MyGraph.cpp:63: warning: unused variable 'edgeNodes' I have figured out that this probably means I'm misusing const somewhere, but I can't figure out where for the life of me. Any information would be appreciated. Thanks!

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  • Converting a const char* into a double

    - by Koning Baard
    I am trying to convert a const char* to a double precision floating point number: int main(const int argc, const char *argv[]) { int i; double numbers[argc - 1]; for(i = 1; i < argc, i += 1) { /* -- Convert each argv into a double and put it in `number` */ } /* ... */ return 0; } Can anyone help me? Thanks

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  • How to push_back without operator=() for const members?

    - by WilliamKF
    How to push_back() to a C++ std::vector without using operator=() for which the default definition violates having const members? struct Item { Item(int value) : _value(value) { } const int _value; } vector<Item> items; items.push_back(Item(3)); I'd like to keep the _value const since it should not change after the object is constructed, so the question is how do I initialize my vector with elements without invoking operator=()?

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  • const keyword in Objective-c

    - by user392412
    int main(int argc, char *argv[]) { @autoreleasepool { const int x = 1; const NSMutableArray *array1 = [NSMutableArray array]; const NSMutableString *str1 = @"1"; NSString * const str2 = @"2"; // x = 2; compile error [array1 addObject:@"2"]; // ok // [str1 appendString:@"2"]; // runtime error // Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Attempt to mutate immutable object with appendString:' // str2 = @"3"; compile error } } my Question is Why array1 addObject is legal and why str1 appendString is forbidden?

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  • Template neglects const (why?)

    - by Gabriel
    Does somebody know, why this compiles?? template< typename TBufferTypeFront, typename TBufferTypeBack = TBufferTypeFront> class FrontBackBuffer{ public: FrontBackBuffer( const TBufferTypeFront front, const TBufferTypeBack back): ////const reference assigned to reference??? m_Front(front), m_Back(back) { }; ~FrontBackBuffer() {}; TBufferTypeFront m_Front; ///< The front buffer TBufferTypeBack m_Back; ///< The back buffer }; int main(){ int b; int a; FrontBackBuffer<int&,int&> buffer(a,b); // buffer.m_Back = 33; buffer.m_Front = 55; } I compile with GCC 4.4. Why does it even let me compile this? Shouldn't there be an error that I cannot assign a const reference to a non-const reference?

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  • Const-Qualification of Main's Parameters in C++

    - by pt2cv
    The C++ standard mandates that all conforming implementations support the following two signatures for main: int main(); int main(int, char*[]); In case of the latter signature, would the addition of (top-level) const-ness break any language rules? For example: int main(const int argc, char** const argv); From my understanding, top-level const qualification doesn't affect the function's signature hash, so it should be legal as far as the specification is concerned. Also, did anyone ever encounter an implementation which rejected this type of modification?

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