Trying to using a template, where the argument is
template<class T, boost::function<long (T*)> &f>
static long myFunc(const vector<boost::shared_ptr<T>> &inputVector)
{ // do stuff}
This is the call I make
long i = myFunc<MyClass, boost::bind(&MyClass::myClassFunc, _1)>(myInputVector);
where the signature of the function is
long myClassFunc() const { return m_value; }
Getting the following compiler error:
error C2975: 'f' : invalid template argument for 'myFunc', expected compile-time constant expression
What do I need to get this to compile?
Binding arguments to a function is a run-time operation. When you pass a value as a template parameter, the value has to be known at compile time. Pass the boost::function as an argument.
template<class T>
static long myFunc(const vector<boost::shared_ptr<T>> &inputVector, boost::function<long (T*)> &f)
{ // do stuff
}
call it like that:
long i = myFunc<MyClass)>(myInputVector, boost::bind(&MyClass::myClassFunc, _1));
Related
In an attempt to call a function template which accepts a type and a parameter/argument of that type as the template parameters/arguments, compiler gives an error which is not produced with similar parameters/arguments. So I was wondering what is the correct parameters/arguments in case of calling the function templates for the member function "operator[]const" of a vector class!
Consider this piece of code:
class test_class{
public:
int member;
int& operator[](size_t) {return member;}
const int& operator[](size_t) const{return member;}
};
typedef std::vector<int> vector_type;
typedef const int&(test_class::* OK_type)(size_t)const;
typedef const int&(vector_type::* not_OK_type)(size_t)const;
static constexpr OK_type OK_pointer = &test_class::operator[];
static constexpr not_OK_type not_OK_pointer = &vector_type::operator[];
template<typename t__, t__>
void function(){}
The above code is alright now consider the main function:
int main() {
function<OK_type, OK_pointer>();
function<not_OK_type, not_OK_pointer>();
return 0;
}
The first call of the function template is OK but not the second one.
The error which compiler produce is:
error: no matching function for call to ‘function<not_OK_type, not_OK_pointer>()’
note: candidate: ‘template<class t__, t__ <anonymous> > void function()’
note: template argument deduction/substitution failed:
error: ‘const int& (std::vector<int>::*)(size_t) const{((const int& (std::vector<int>::*)(size_t) const)std::vector<int>::operator[]), 0}’ is not a valid template argument for type ‘const int& (std::vector<int>::*)(long unsigned int) const’
function<not_OK_type, not_OK_pointer>();
note: it must be a pointer-to-member of the form ‘&X::Y’
Interestingly even if the function template was formed as:
template<auto>
void function(){}
it would cause the same result.
I must add that in the case of non const version, error is the same (for std::vector).
So I am wondering
A: what is wrong?
B: Considering that if there was a mismatch between the not_OK_type and &vector_type::operator[], then compiler would also give an error in case of:
static constexpr not_OK_type not_OK_pointer = &vector_type::operator[];
Is there a difference between types that can be used as constexpr and the type that can be used as a template parameter/argument?
The issue is typedef const int&(vector_type::* not_OK_type)(size_t)const;.
If you see stl_vector.h (here), the operator[] is declared as noexcept at line 1040.
But in the declaration of not_OK_type variable, noexcept is not present. That's why the compiler complains.
For getting rid of compilation error, add noexcept to not_OK_type variable. Like this:
typedef const int&(vector_type::* not_OK_type)(size_t)const noexcept;
Working code:
#include <vector>
class test_class{
public:
int member;
int& operator[](size_t) {return member;}
const int& operator[](size_t) const{return member;}
};
typedef std::vector<int> vector_type;
typedef const int&(test_class::* OK_type)(size_t)const;
typedef const int&(vector_type::* not_OK_type)(size_t)const noexcept;
static constexpr OK_type OK_pointer = &test_class::operator[];
static constexpr not_OK_type not_OK_pointer = &vector_type::operator[];
template<typename t__, t__>
void function(){}
int main() {
function<OK_type, OK_pointer>();
function<not_OK_type, not_OK_pointer>();
return 0;
}
#Kunal Puri, provided a correct answer for the the section A of the question.
As for the section B of the question, I guess the clue can be found in one exception that applies when converted constant expression used as template argument for a non-type template parameter.
According to (https://en.cppreference.com/w/cpp/language/constant_expression), a converted constant expression under certain condition and specifically in case of conversion of pointer to noexcept function to pointer to function is a constant expression. Which explains why compiler did not produce an error in case of:
static constexpr not_OK_type not_OK_pointer = &vector_type::operator[];
However according to (https://en.cppreference.com/w/cpp/language/template_parameters), this type of converted constant expression can not be used as a pointer to non-static data members(and maybe also a pointer to non-static member functions) for a non-type template parameter.
This exception might be the source of conflict between types that can be used as constexpr and as template argument, although statements in the later source are vague and not directly linked to the case.
Why doesn't the following code compile (in C++11 mode)?
#include <vector>
template<typename From, typename To>
void qux(const std::vector<From>&, To (&)(const From&)) { }
struct T { };
void foo(const std::vector<T>& ts) {
qux(ts, [](const T&) { return 42; });
}
The error message is:
prog.cc:9:5: error: no matching function for call to 'qux'
qux(ts, [](const T&) { return 42; });
^~~
prog.cc:4:6: note: candidate template ignored: could not match 'To (const From &)' against '(lambda at prog.cc:9:13)'
void qux(const std::vector<From>&, To (&)(const From&)) { }
^
But it doesn't explain why it couldn't match the parameter.
If I make qux a non-template function, replacing From with T and To with int, it compiles.
A lambda function isn't a normal function. Each lambda has its own type that is not To (&)(const From&) in any case.
A non capturing lambda can decay to To (*)(const From&) in your case using:
qux(ts, +[](const T&) { return 42; });
As noted in the comments, the best you can do to get it out from a lambda is this:
#include <vector>
template<typename From, typename To>
void qux(const std::vector<From>&, To (&)(const From&)) { }
struct T { };
void foo(const std::vector<T>& ts) {
qux(ts, *+[](const T&) { return 42; });
}
int main() {}
Note: I assumed that deducing return type and types of the arguments is mandatory for the real problem. Otherwise you can easily deduce the whole lambda as a generic callable object and use it directly, no need to decay anything.
If you don't need to use the deduced To type, you can just deduce the type of the whole parameter:
template<typename From, typename F>
void qux(const std::vector<From>&, const F&) { }
Correct me if I am wrong, but template parameters deduction deduces only exact types without considering possible conversions.
As a result the compiler cannot deduce To and From for To (&)(const From&) because qux expects a reference to function, but you provide a lambda which has its own type.
You have left absolutely no chance to compiler to guess what is To. Thus, you need to specify it explicitly.
Also, lambda here needs to be passed by pointer.
Finally, this version compiles ok:
template<typename From, typename To>
void qux(const std::vector<From>&, To (*)(const From&)) { }
struct T { };
void foo(const std::vector<T>& ts) {
qux<T,int>(ts,[](const T&) { return 42; });
}
You're expecting both implicit type conversions (from unnamed function object type to function reference type) and template type deduction to happen. However, you can't have both, as you need to know the target type to find the suitable conversion sequence.
But it doesn't explain why it couldn't match the parameter.
Template deduction tries to match the types exactly. If the types cannot be deduced, deduction fails. Conversions are never considered.
In this expression:
qux(ts, [](const T&) { return 42; });
The type of the lambda expression is some unique, unnamed type. Whatever that type is, it is definitely not To(const From&) - so deduction fails.
If I make qux a non-template function, replacing From with T and To with int, it compiles.
That is not true. However, if the argument was a pointer to function rather than a reference to function, then it would be. This is because a lambda with no capture is implicitly convertible to the equivalent function pointer type. This conversion is allowed outside of the context of deduction.
template <class From, class To>
void func_tmpl(From(*)(To) ) { }
void func_normal(int(*)(int ) ) { }
func_tmpl([](int i){return i; }); // error
func_tmpl(+[](int i){return i; }); // ok, we force the conversion ourselves,
// the type of this expression can be deduced
func_normal([](int i){return i; }); // ok, implicit conversion
This is the same reason why this fails:
template <class T> void foo(std::function<T()> );
foo([]{ return 42; }); // error, this lambda is NOT a function<T()>
But this succeeds:
void bar(std::function<int()> );
bar([]{ return 42; }); // ok, this lambda is convertible to function<int()>
The preferred approach would be to deduce the type of the callable and pick out the result using std::result_of:
template <class From,
class F&&,
class To = std::result_of_t<F&&(From const&)>>
void qux(std::vector<From> const&, F&& );
Now you can pass your lambda, or function, or function object just fine.
I have created a hasher class for a custom type I'm using, but it has a constructor that takes an argument. I can't figure out the syntax to use this in an unordered_set.
class Hasher {
unsigned arg;
public:
Hasher(unsigned a) : arg(a) {}
size_t operator()(const MyType& t) const {
return calculate_hash(arg, t);
}
}
int main() {
unordered_set<MyType, Hasher(2)> myset; // compilation error
}
The error message:
In file included from Tetrahedron.cc:5:
./Triangulation.h:52:29: error: template argument for template type parameter must be a type
unordered_set<TetraFace,FaceHasher(2)> faces2;
^~~~~~~~~~~~~
/bin/../lib/gcc/x86_64-redhat-linux/6.3.1/../../../../include/c++/6.3.1/bits/unordered_set.h:90:11: note: template parameter is declared here
class _Hash = hash<_Value>,
^
I also tried
unordered_set<MyType, Hasher> myset(Hasher(2));
but I still get an error:
In file included from Tetrahedron.cc:5:
./Triangulation.h:52:59: error: expected ')'
unordered_set<TetraFace,FaceHasher> faces2(FaceHasher(2));
^
./Triangulation.h:52:58: note: to match this '('
unordered_set<TetraFace,FaceHasher> faces2(FaceHasher(2));
^
You're getting a compile error there because you're trying to pass an object (i.e. instance) of type Hasher as a template argument.
Like your error describes: template argument for template type parameter must be a type
It's expecting a type, and you're passing in a value.
Parameterize the arg at the type level.
template<unsigned A>
class Hasher {
unsigned arg = A;
public:
size_t operator()(const int& t) const {
std::cout << arg << std::endl;
return 0;
}
};
int main() {
std::unordered_set<int, Hasher<2>> myset;
myset.insert(5); // prints 2
std::unordered_set<int, Hasher<3>> myset2;
myset2.insert(3); // prints 3
}
Unfortunately it is not possible to construct a std::unorderd_set with just the hash object. All of the constructors that take the hash object have a parameter before it for bucket_count. You would need to specify the value for it like
unordered_set<MyType, Hasher> myset(some_bucket_count_value, Hasher(2));
If you do not want to do that then you have to make Hasher default constructable.
Also not that
return calculate_hash(arg);
Is not going to work as you will always hash arg no matter what MyType you pass. You need to be hashing the MyType object for the std::unordered_set to really work.
I've got some template code that I'm modifying and I've run into an odd error that I can't work around. I was able to recreate the problem with the below simpler (but admittedly pointless) code snippet:
struct Widget
{
};
template <typename A>
class Foo
{
public:
template <int numA>
inline bool funcCall()
{
return numA > 0;
}
inline bool funcCallNoTemplate()
{
return false;
}
};
template <typename B>
class Bar : public Foo<B>
{
public:
// doesn't work
bool concrete()
{
return Foo<B>::funcCall<5>();
}
// works fine
bool other()
{
return Foo<B>::funcCallNoTemplate();
}
};
int main()
{
Bar<Widget> b;
b.concrete();
b.other();
return 0;
}
The error I get with GCC 4.7 is the following (line 30 is the body of Bar::concrete):
example.cxx: In member function ‘bool Bar<B>::concrete()’:
example.cxx:30: error: expected primary-expression before ‘)’ token
example.cxx: In member function ‘bool Bar<B>::concrete() [with B = Widget]’:
example.cxx:43: instantiated from here
example.cxx:30: error: invalid operands of types ‘<unresolved overloaded function type>’ and ‘int’ to binary ‘operator<’
It seems like the compiler can't even parse this correctly, am I missing something here that makes that line completely bogus?
It seems like the compiler can't even parse this correctly, am I missing something here that makes that line completely bogus?
Yes. You need to use the template disambiguator:
return Foo<B>::template funcCall<5>();
// ^^^^^^^^
This way you will tell the compiler to parse the dependent name funcCall as the name of a member function template, and the subsequent angular brackets as delimiters for the corresponding template arguments.
Without it, funcCall will be parsed as the name of a data member, while < and > will be parsed as less-than and greater-than, respectively.
I am learning c++ template concepts. I do not understand the following.
#include <iostream>
#include <typeinfo>
using namespace std;
template <typename T>
T fun(T& x)
{
cout <<" X is "<<x;
cout <<"Type id is "<<typeid(x).name()<<endl;
}
int main ( int argc, char ** argv)
{
int a[100];
fun (a);
}
What i am trying?
1) T fun (T & x)
Here x is a reference, and hence will not decayed 'a' into pointer type,
but while compiling , i am getting the following error.
error: no matching function for call to ‘fun(int [100])’
When I try non-reference, it works fine. As I understand it the array is decayed into pointer type.
C-style arrays are very basic constructs which are not assignable, copyable or referenceable in the way built-ins or user defined types are. To achieve the equivalent of passing an array by reference, you need the following syntax:
// non-const version
template <typename T, size_t N>
void fun( T (&x)[N] ) { ... }
// const version
template <typename T, size_t N>
void fun( const T (&x)[N] ) { ... }
Note that here the size of the array is also a template parameter to allow the function to work will all array sizes, since T[M] and T[N] are not the same type for different M, N. Also note that the function returns void. There is no way of returning an array by value, since the array is not copyable, as already mentioned.
The problem is in the return type: you cannot return an array because arrays are non-copiable. And by the way, you are returning nothing!
Try instead:
template <typename T>
void fun(T& x) // <--- note the void
{
cout <<" X is "<<x;
cout <<"Type id is "<<typeid(x).name()<<endl;
}
And it will work as expected.
NOTE: the original full error message (with gcc 4.8) is actually:
test.cpp: In function ‘int main(int, char**)’:
test.cpp:17:10: error: no matching function for call to ‘fun(int [100])’
fun (a);
^
test.cpp:17:10: note: candidate is:
test.cpp:7:3: note: template<class T> T fun(T&)
T fun(T& x)
^
test.cpp:7:3: note: template argument deduction/substitution failed:
test.cpp: In substitution of ‘template<class T> T fun(T&) [with T = int [100]]’:
test.cpp:17:10: required from here
test.cpp:7:3: error: function returning an array
The most relevant line is the last one.