#include <iostream>
template <typename T1, typename T2>
bool func(const T1& t, const T2& t2) {
return t == t2;
}
class Base {
public:
bool operator ==(const Base&) const { return true;}
Base(int y) : x(y) {}
operator int() {
return x;
}
int x;
};
int main() {
func<long, Base>(4L, Base(5)); // not ok
func<long, long>(4L, Base(5)); //ok
}
Can somebody elaborate why the first version does not work? In otherwords why does the binary operator == in func not use the conversion operator int to convert the template parameter bound to Base into int?
Is there anyway to make version 1 work by only modifying the class Base?
Your func accepts its parameters by const reference, but the operator int() defined in your Base class is a non-const member function. Mark it as a const member function, as shown below, and your code will compile:
#include <iostream>
template <typename T1, typename T2>
bool func(const T1& t, const T2& t2) {
return t == t2;
}
class Base {
public:
bool operator ==(const Base&) const { return true;}
Base(int y) : x(y) {}
operator int() const {
return x;
}
int x;
};
int main() {
func<long, Base>(4L, Base(5)); // ok now!
func<long, long>(4L, Base(5)); // ok
}
Live example
Related
A simple C++ OO question regrading templates and operator overloading: In the following class, I have overloaded the index operator twice:
template<class A, class B>
class test
{
A a1;
B a2;
public:
A& operator[](const B&);
B& operator[](const A&);
};
Now, if I instantiate an object of this template class with the same typenames:
test<int, int> obj;
calling the index operator will result in an error, because the two overloaded functions will have the same signatures.
Is there any way to resolve this issue?
Sorry, if this is a basic question. I am still learning!
You can add a partial specialization:
template<class A>
class test<A, A>
{
A a1, a2;
public:
A& operator[](const A&);
};
You can avoid this issue and make the code more robust and expressive by converting the index to some other type that clarifies what the user wants. Usage would be like this:
bidirectional_map<int, int> myTest;
int& valueFor1 = myTest[Key{1}];
int& key1 = myTest[Value{valueFor1}];
Implemented like this:
template<class TKey>
struct Key { const TKey& key; };
template<class TValue>
struct Value { const TValue& value; };
// Deduction guides (C++17), or use helper functions.
template<class TValue>
Value(const TValue&) -> Value<TValue>;
template<class TKey>
Key(const TKey&) -> Key<TKey>;
template<class TKey, class TValue>
class bidirectional_map
{
TKey a1; // Probably arrays
TValue a2; // or so?
public:
TValue & operator[](Key<TKey> keyTag) { const TKey & key = keyTag.key; /* ... */ }
TKey & operator[](Value<TValue> valueTag) { const TValue& value = valueTag.value; /* ... */ }
};
Now, Key and Value are popular names so having them "taken up" by these auxiliary functions is not the best. Also, this is all just a pretty theoretical exercise, because member functions are of course a much better fit for this task:
template<class TKey, class TValue>
class bidirectional_map
{
TKey a1; // Probably arrays
TValue a2; // or so?
public:
TValue& getValueForKey(const TKey& key) { /* ... */ }
TKey& getKeyForValue(const TValue& value) { /* ... */ }
};
In C++2a, you might use requires to "discard" the function in some case:
template<class A, class B>
class test
{
A a1;
B a2;
public:
A& operator[](const B&);
B& operator[](const A&) requires (!std::is_same<A, B>::value);
};
Demo
Here is an example solution using if constexpr that requires C++17:
#include <type_traits>
#include <cassert>
#include <string>
template <class A, class B>
class test
{
A a1_;
B b1_;
public:
template<typename T>
T& operator[](const T& t)
{
constexpr bool AequalsB = std::is_same<A,B>();
constexpr bool TequalsA = std::is_same<T,A>();
if constexpr (AequalsB)
{
if constexpr (TequalsA)
return a1_; // Can also be b1_, same types;
static_assert(TequalsA, "If A=B, then T=A=B, otherwise type T is not available.");
}
if constexpr (! AequalsB)
{
constexpr bool TequalsB = std::is_same<T,B>();
if constexpr (TequalsA)
return a1_;
if constexpr (TequalsB)
return b1_;
static_assert((TequalsA || TequalsB), "If A!=B, then T=A || T=B, otherwise type T is not available.");
}
}
};
using namespace std;
int main()
{
int x = 0;
double y = 3.14;
string s = "whatever";
test<int, int> o;
o[x];
//o[y]; // Fails, as expected.
//o[s]; // Fails, as expected
test<double, int> t;
t[x];
t[y];
//t[s]; // Fails, as expected.
return 0;
};
Problem Description and Question
I have a template class Class1. It contains in map in which I want to insert structures A or B.
The problem is that the structures A and B have different types of member variables. Structure A has an std::string member variable whereas structure B has an int member variable.
The comparator is based on structure A. So obviously when I want to insert a structure B it will not compile.
Class1<B,B> c2;
c2.AddElement({1},{1});
How can I fix that design Issue? For instance is it possible to keep Class1 as template class and do something to TestCompare?
I also have a constraint. I cannot modify the structures A and B. they are written in C code. I have no right to change them because they are external codes used by other users. I just simplified the code as much as possible.
Source Code
The code was compiled on cpp.sh
#include <iostream>
#include <string>
#include <map>
typedef struct {
std::string a;
} A;
typedef struct {
int b;
} B;
template<typename T1, typename T2> class Class1 {
public :
struct TestCompare {
bool operator()(const T1 & lhs, const T1 & rhs) const {
return lhs.a < rhs.a;
}
};
Class1() {}
~Class1() {}
void AddElement(const T1 & key, const T2 & value) {
m.emplace(key, value);
}
private :
std::map<T1,T2,TestCompare> m;
};
int main()
{
Class1<A,A> c1;
c1.AddElement({"1"},{"1"});
// Problem here. Obviously it will not compile because the Operator is using
// the member variable of struct A.
//Class1<B,B> c2;
//c2.AddElement({1},{1});
//return 0;
}
New Source code
// Example program
#include <iostream>
#include <string>
#include <map>
typedef struct {
std::string a;
} A;
typedef struct {
int b;
} B;
bool operator<(const A & lhs, const A & rhs) {
return lhs.a < rhs.a;
}
bool operator<(const B & lhs, const B & rhs) {
return lhs.b < rhs.b;
}
template<typename T1, typename T2> class Class1 {
public :
Class1() {}
~Class1() {}
void AddElement(const T1 & key, const T2 value) {
m.emplace(key, value);
}
std::map<T1,T2> getMap() {
return m;
}
private :
std::map<T1,T2> m;
};
int main()
{
Class1<A,A> c1;
c1.AddElement({"1"},{"1"});
// Problem here. Obviously it will not compile because the Operator is using
// the member variable of struct A.
Class1<B,B> c2;
c2.AddElement({1},{1});
c2.AddElement({2},{2});
for(const auto &e: c2.getMap()) {
std::cout << e.first.b << " " << e.first.b << std::endl;
}
return 0;
}
I guess you could remove TestCompare from Class1 and template that.
template<typename T> struct TestCompare {
bool operator()(const T & lhs, const T & rhs) const {
// default implementation
return lhs < rhs;
}
};
template<typename T1, typename T2> class Class1 {
...
private :
std::map<T1,T2,TestCompare<T1>> m;
}
You could then specialise TestCompare for A and B
template<> struct TestCompare<A> {
bool operator()(const A & lhs, const A & rhs) const {
return lhs.a < rhs.a;
}
};
template<> struct TestCompare<B> {
bool operator()(const B & lhs, const B & rhs) const {
return lhs.b < rhs.b;
}
};
EDIT:
Actually you could just use std::less instead of TestCompare. It amounts to pretty much the same thing, and std::map uses std::less by default.
TestCompare requires that every type you use must have a member a that can be compared using <. That's a lot of requirements, which implies a terrible design. Add a 3rd template parameter that will be used to pass a function or a functor that compares the objects
struct CompareA {
bool operator()(A const & lhs, A const & rhs) const {
return lhs.a < rhs.a;
}
};
struct CompareB {
bool operator()(B const& lhs, B const& rhs) const {
/*...*/
}
};
template<typename KeyT, typename ValueT, typename Compare> class Dict {
public :
Class1() {}
~Class1() {}
void AddElement(KeyT const & key, ValueT const & value) {
m.emplace(key, value);
}
private :
std::map<KeyT, ValueT, Compare> m;
};
Dict<A, B, CompareA> dictA;
Dict<B, B CompareB> dictB;
You could specialize the struct TestCompare, like john has suggested in his answer, and provide it as the default template argument
template<typename KeyT, typename ValueT, typename Compare = TestCompare<KeyT>> class Dict { /*...*/ };
Such solution will allow you to provide only 2 arguments, like so
Dict<B, B> dict;
while still maintaining the ability to provide another comparer if necessary.
I am trying to wrap some templated functions into some binary functors like below. When I try to compile the code I have the error
error: no match for call to ‘(QtyAsc) (myobj&, myobj&)
I thought that being operator() in QtyAsc a function in a class, the template deduction mechanism would have worked but it seems that the compiler doesn't accept myobj classes as valid types for it.
Is it maybe because of the call to boost::bind? I was trying to provide a default implementation for the second templated argument (unfortunately I cannot use C++11 with default templated arguments).
class myobj {
public:
myobj(int val) : qty_(val) {}
int qty() { return qty_;}
private:
int qty_;
};
template<class T>
int get_quantity(const T& o) {
throw runtime_error("get_quantity<T> not implemented");
}
template<>
int get_quantity(const myobj& o) {
return o.qty();
}
struct QtyAsc {
template<class T, class QEx >
bool operator()(const T& o1, const T& o2, QEx extr = boost::bind(&get_quantity<T>,_1)) const {
if(extr(o1) < extr(o2))
return true;
return false;
}
};
int main() {
myobj t1(10),t2(20);
QtyAsc asc;
if(asc(t1,t2))
cout << "Yes" << endl;
}
If you can't use C++11, just provide an additional overload:
struct QtyAsc {
template<class T, class QEx >
bool operator()(const T& o1, const T& o2, QEx extr) const {
return extr(o1) < extr(o2);
}
template<class T>
bool operator()(const T& o1, const T& o2) const {
return operator()(o1, o2, &get_quantity<T>);
}
};
(I've omitted the unnecessary boost::bind.) Also, you will need to declare myobj::qty to be const:
int qty() const {
return qty_;
}
since you want to invoke it on const objects. (Live demo)
For example:
struct B{};
struct A {
const B& findB() const { /* some non trivial code */ }
// B& findB() { /* the same non trivial code */ }
B& findB() {
const A& a = *this;
const B& b = a.findB();
return const_cast<B&>(b);
}
};
The thing is I want to avoid repeating the same logic inside the constant findB and non-constant findB member function.
Yes, you can cast the object to const, call the const version, then cast the result to non-const:
return const_cast<B&>(static_cast<const A*>(this)->findB());
Casting away const is safe only when the object in question was not originally declared const. Since you are in a non-const member function, you can know this to be the case, but it depends on the implementation. Consider:
class A {
public:
A(int value) : value(value) {}
// Safe: const int -> const int&
const int& get() const {
return value;
}
// Clearly unsafe: const int -> int&
int& get() {
return const_cast<int&>(static_cast<const A*>(this)->get());
}
private:
const int value;
};
Generally speaking, my member functions are short, so the repetition is tolerable. You can sometimes factor the implementation into a private template member function and call that from both versions.
I think, that using cast here is ok, but if you definitely want to avoid it, you can use some template magic:
struct B
{
B(const B&)
{
std::cout << "oops I copied";
}
B(){}
};
struct A {
public:
A(){}
A(const A&){ std::cout << "a is copied:(\n";}
const B& findB() const { return getter(*this); }
B& findB() { return getter(*this); }
private:
template <typename T, typename V>
struct same_const
{
typedef V& type;
};
template <typename T, typename V>
struct same_const<const T, V>
{
typedef const V& type;
};
template <typename T>
static typename same_const<T,B>::type getter(T& t) { return t.b;}
B b;
};
int main()
{
A a;
const A a_const;
const B& b1 = a.findB();
B& b2 = a.findB();
const B& b3 = a_const.findB();
//B& b4 = a_const.findB();
}
I wrote a class and I wanted to implement an iterator for it ( as shown in the following code ). I needed to overload a variety of operators and I faced the error mentioned below:
class BaseClass
{
virtual ~BaseClass() {}
};
template<class T>
class AbstractBaseOrgan: public BaseClass
{
public:
typedef T value;
template<class TT>
class AbstractBaseIterator:
public std::iterator<std::random_access_iterator_tag,
typename std::iterator_traits<TT>::value_type>
{
protected:
TT _M_current;
const TT&
base() const
{ return this->_M_current; }
};
protected:
value te;
};
template<typename Iter>
inline bool
operator<(const typename AbstractBaseOrgan<typename
std::iterator_traits<Iter>::value_type>::template
AbstractBaseIterator<Iter>& lhs,
const typename AbstractBaseOrgan<typename
std::iterator_traits<Iter>::value_type>::template
AbstractBaseIterator<Iter>& rhs)
{ return lhs.base() < rhs.base(); }
int main()
{
AbstractBaseOrgan<int>::AbstractBaseIterator<int*> temp;
AbstractBaseOrgan<int>::AbstractBaseIterator<int*> temp2;
int ttemp;
if(operator< (temp,temp2))
ttemp = 0;
return 0;
}
Compiling it gives me the following error:
error: no matching function for call to ‘operator<(AbstractBaseOrgan<int>::AbstractBaseIterator<int*>&, AbstractBaseOrgan<int>::AbstractBaseIterator<int*>&)’
Any idea what might cause this?
4 In most cases, the types, templates, and non-type values that are
used to compose P participate in template argument deduction. That is,
they may be used to determine the value of a template argument, and
the value so determined must be consistent with the values determined
elsewhere. In certain contexts, however, the value does not
participate in type deduction, but instead uses the values of template
arguments that were either deduced elsewhere or explicitly specified.
If a template parameter is used only in non-deduced contexts and is
not explicitly specified, template argument deduction fails.
The non-deduced contexts are:
— The nested-name-specifier of a type that was specified using a qualified-id.
You can avoid this by few ways. First way - make operator < friend for class AbstractIteratorBase, or its member.
template<class TT>
class AbstractBaseIterator:
public std::iterator<std::random_access_iterator_tag,
typename std::iterator_traits<TT>::value_type>
{
public:
template<typename Iter>
friend bool operator < (const AbstractBaseIterator<Iter>& lhs, const AbstractBaseIterator<Iter>& rhs)
{
return lhs.base() < rhs.base();
}
protected:
TT _M_current;
const TT&
base() const
{ return this->_M_current; }
};
Second variant is define AbstractBaseIterator class not in template class. And then typedef AbstractBaseIterator<T> iterator; in AbstractBaseOrgan. If you can use C++11 you can use something like this.
class BaseClass
{
virtual ~BaseClass() {}
};
template<class TT>
class AbstractBaseIterator:
public std::iterator<std::random_access_iterator_tag,
typename std::iterator_traits<TT>::value_type>
{
protected:
TT _M_current;
const TT&
base() const
{ return this->_M_current; }
};
template<typename Iter>
bool operator < (const AbstractBaseIterator<Iter>& lhs, const AbstractBaseIterator<Iter>& rhs)
{
return lhs.base() < rhs.base();
}
template<class T>
class AbstractBaseOrgan: public BaseClass
{
public:
typedef T value;
template<typename TT>
using iterator = AbstractBaseIterator<TT>;
protected:
value te;
};
int main()
{
AbstractBaseOrgan<int>::iterator<int*> temp;
AbstractBaseOrgan<int>::iterator<int*> temp2;
int ttemp;
if(operator< (temp,temp2))
ttemp = 0;
return 0;
}