Is this really the preferred way to get the return type, for use in a derived class, of a function defined in the template parameter?
template<class PARAMETER> class C {
protected:
typedef typeof (reinterpret_cast<PARAMETER*>(0))->function() returntype;
};
This works just fine for me, but seems inelegant.
I have this code:
#include <iostream>
#include <vector>
#include <map>
#include <string>
using namespace std;
[Code]...
and it does not compile.
The error is:
test.cpp: In function ‘int main()’:
test.cpp:20:30: error: no matching function for call to ‘func1(std::vector<int>&)’
test.cpp:20:30: note: candidate is:
test.cpp:8:45: note: template<class T, class U> std::map<T, T> func1(U)
test.cpp:8:45: note: template argument deduction/substitution failed:
test.cpp:20:30: note: couldn't deduce template parameter ‘T’
I have a class where a method based on the type passed I should return a value.
prototype declared in the header file:
template <typename T>int getNum() const;
Code of the cpp file:
template <typename T> int class::getNum() const{
int c = 0;
for(int i=0;i<v.size();i++)
if(typeid(*(Pro*)v.at(i)) == typeid(T)) c++;
return c;
}
To invoke the method as I do:
ostream & operator << (ostream & os, Pro & obj) {
return os << obj.getNum();
}
Pseudocode:
template<typename T /*, some parameter for member_function */>
class Foo {
public:
void someFunction() {
T t;
t.member_fuction(...);
} }
I'm trying to make the call to T::member_function a templated value because member_function might vary by name in my scenario. Since std::mem_fn isn't a 'type', i can't do something like Foo<std::string, std::mem_fn(&std::string::clear)> foo;
I also want to take into account that member_function might have more than one parameter. That is, the first parameter will always be something known but there might be other defaulted parameters.
The only thing I can think of is to make a proxy structure, something like this:
template<typename T, T> struct proxy;
template<typename T, typename R, typename... Args, R (T::*member_function)(Args...)>
struct proxy<R (T::*)(Args...), member_function> {
R operator()(T &obj, Args.. args) {
return (obj.*member_function)(std::forward<Args>(args)...);
} }
Which would then allow me to do (for example) this:
Foo<std::string, proxy<void(std::string::*)(), &std::string::clear> >
when Foo is implemented like this:
template<typename T, typename member_function>
class Foo {
public:
void someFunction() {
T t;
member_function()(t);
} };
That implementation works for me.
Is there any function which can return parameter list of a function.
For example : get_param(f(int x,char y )) return parameter x-> int y-> char
and f_name ->f
Due to the nature of this requirement, I've made a very minimal example, which would adequately solve my issue, without resorting to use of pointers or copy constructors.
Basically I'm trying to pass an object as a reference to the template function, rather than a copy as it's seeing. I'm needing to do this without editing Obj::Call to accommodate a reference as its first parameter, as it'd break other calls.
You'll notice in the following code the object will be destroyed upon passing, while the object defined is still in-scope due to the infinite end loop.
#include <iostream>
#include <string>
using namespace std;
class Obj {
public:
string name;
[Code] ....
In the past I tried ref(), which appeared to stop this happening, however it created a blank copy of the object instead.
void extf(int a) { }
template<typename P>
struct A {
// 1
template< void (*F)(P) >
static void call(P arg) {
[Code]...
Why it is not working? What would be a proper way to pass function pointer as a template parameter?
Let me put it into the code snippet:
/**
This class build the singleton design pattern.
Here you have full control over construction and deconstruction of the object.
*/
template<class T>
class Singleton
[Code]....
I am getting error at the assertion points when i call to the class as follows:
osgOpenCL::Context *cxt = osgOpenCL::Singleton<osgOpenCL::Context>::getPtr();
I tried commenting assertion statements and then the debugger just exits at the point where getPtr() is called.
error says "cannot convert 'int*' to 'int' in function main()
and also
type mismatch in parameter in function sort(int,int)
Heres the code:
#include<iostream.h>
#include<conio.h>
void main() {
void sort(int,int);
clrscr();
[Code] .....
Basically I'm trying to pass an object as a reference to the template function, rather than a copy as it's seeing. I'm needing to do this without editing Obj::Call to accommodate a reference as its first parameter, as it'd break other calls.
You'll notice in the following code the object will be destroyed upon passing, while the object defined is still in-scope due to the infinite end loop.
#include <iostream>
#include <string>
using namespace std;
class Obj {
public:
string name;
Obj(string name): name(name) {cout << "create " << this << endl;}
[code]....
In the past I tried ref(), which appeared to stop this happening, however it created a blank copy of the object instead.
Say I have overloaded functions with different arguments AND return types. Each set of argument has only one corresponding return type.
Code:
Vector grad(Scalar)
Tensor grad(Vector)
Later I have:
Code:
template <class T>
void test(T x) {
... Y = grad(x)
}
Then how do I automatically declare the type of Y. Do I need to make the grad function a template and specialize each of them ?
i heard that it was int but later it is deprecated , so at present what is the default return type of a function ?
View 4 Replies View RelatedI want my function to return the type mpz_t, but I'm not sure how?
I've tried: mpz_t MyFunction(mpz_t A, mpz_t B){}
But it didn't work, here is my code so far, I have bolded the parts of the code which are causing errors and added the errors in the comments:
Code:
#include <iostream>
#include <mpir.h>
using namespace std;
???? A(mpz_t m, mpz_t n){
mpz_t mmo; //used to store the value of m, minus one
[Code] .....
Is there a way around this?
I want to detect the type in a function template, like this:
template <class myType> myType Function (myType a, myType b) {
//Detect the myType
If (myType is int)
[Code] ......
Is that possible?
how I want the code to look. Only problem is it doesn't work (Line 11). I have some experience with templates but I'm not a pro.
Basically I want the "Channels<3>" to be a type that I can use to specify a Cable with similar to vector<float/int> it would be Cable<Channels<2 or 3>>.
What have I messed up with the syntax?
#include <iostream>
#include <vector>
using namespace std;
[Code].....
I am having problems with my function definition of a function that should return a structure value.
This is the error I get compute.cpp(9): error C2146: syntax error : missing ';' before identifier 's_advertisebus'
The error is on the line where I start my function definition typing my function type as a structure. A long time ago in c the keyword struct is used with the structure type like struct s_advertisebus s_readadbus(). I tried it both ways but I got errors.
// struct.h
#ifndef STRUCT_H
#define STRUCT_H
struct s_advertisebus {
int nnumberofads;
float fpercentused;
[Code] ....
I always have confusions while using pointers with functions both as arguments and as return type.
View 1 Replies View RelatedIf we are using strcpy() for copying the string. As we are passing pointers to it It will copy the string & no need to return the string .This function will finely work with return type as void then why Ritchie has used it as char* strcpy()?
View 4 Replies View RelatedI have a class "Result" with a single template function Set(const std::string& arName, T& val) and a specialization of this function Set<Real>(const std::string& arName, Real& val) where Real is a typedef for double. The class is in a shared library and I use it in my main program. If I do result->Set<GLOBAL::Real>("U", 100.0); the wrong template function is called!
I check this by the output with std::cout.
Maybe it's a problem with the typedef.
If I link the object file of the Result class directly to my main program (no shared library), it works.
typedefs.hpp:
namespace GLOBAL {
typedef double Real;
} results.hpp
#include <iostream>
[Code] ....
Unless I'm missing something, it's now possible(ish)? A little concept is below, very rough around the edges. Still though, if this is valid by standard C++, why can't we have built-in support for float / double template parameters?
#include <iostream>
#include <tuple>
template<int Numerator, int Denominator>
struct Float {
constexpr static float value()
[Code] ....
My errors are at the end of the program in two function calls within the definition of the InsertByValue function. g++ does not seem to recognize NumArray as a valid parameter.
#include <iostream>
#include <assert.h>
using namespace std;
const int CAPACITY = 20;
/* Displays the content of an int array, both the array and the size of array will be passed as parameters to the function
@param array: gives the array to be displayed
@param array_size: gives the number of elements in the array */
void DisplayArray (int array[], int array_size);
[Code] ....
I have the defined a class Seconds with a template<int,int> constructor. Unfortunately, I can't seem to get an instance. Here's the class :
class Seconds {
public :
template <int num, int den> Seconds(long int);
}
And when I try to instantiate it :
Seconds *millis = new Seconds<1,1000>(30); // template argument deduction/substitution failed:
// couldn't deduce template parameter ‘num’
Is it not the right way to call the constructor?
I have a triple hierarchy class:
template<class T> class Singleton;
class Base;
class Sub : public Base, public Singleton<Sub>;
I' using underlying auto pointers, that's why Singleton is a template class and Sub passes itself as a template parameter. I'm developing Singleton and Base and a public API allows anyone to add their own sub classes. I actually want a real triple hierarchy like this:
template<class T> class Singleton;
class Base : public Singleton<Base>;
class Sub : public Base;
So that external developers don't have to worry about templates and complexity. The problem with this is that my implementation in Singleton will now call the constructor of Base whenever I create an instance of Sub (since the template parameter is Base).I was wondering if this could be done by pre-processor macros:
template<class T> class Singleton;
class Base : public Singleton<__CLASS_NAME__>;
class Sub : public Base;
Where __CLASS_NAME__ is the class name that will be replaced by the pre-processor. Theoretically this should be possible, since the __PRETTY_ FUNCTION__ macro actually returns the class name. The problem is that one cannot do string-manipulation to remove the function name from __PRETTY_FUNCTION__.
how I can accomplish this so that the Sub class is not aware of inheriting from a Singleton<template> class?
I've been trying to create a templated class that takes a template as a parameter. I'd like to specialise this class for certain partial specializations of the template parameter but can't seem to figure out how to do it nor find anything online, (although I may be searching for the wrong thing).
As an example, say I have a class A that takes a template class with two parameters as its parameter:
template< template<class X, class Y> class Z > class A {};
I'd like to have a general version of A, for a general version of Z, but a specialisation of A for a specialisation of Z, e.g. where X is int but Y is still any type.
I've been playing around with this piece of code:
#include <iostream>
#include <string>
#include <algorithm>
template <void(*funky)(const std::string&)>
void callback()
{funky("Hello World!");}
[Code] ....
But when I try to build it, I get this error on line 24:could not convert template argument 'lambda' to 'void (*)(const string&) {aka void (*)(const std::basic_string<char>&)}'|
I thought the lambda expression I wrote would decay to a function pointer matching the template parameter. I can guess that the constexpr qualifier might have changed the type, but without it my compiler complains that lambda needs to be declared as constexpr...
So is there a way to pass lambda expressions as template parameters?
Without having to use std::function
Is there an "easy" way to define a template with a template parameter that could resolve to "nothing".
Code:
template <typename T1, typename T2, typename T3>
struct one_to_three {
public:
T1 t1;
T2 t2;
T3 t3;
};
Now I want to be able to instantiate that in such a way that T3 and T2 (if T3 also) resolve to "nothing".
Don't point me at tuple<>, this has to be a single structure. No dummy's, nothing effectively means nothing.
so on Win32
Code:
one_to_three <int, int, int> x;
sizeof(x) == 12;
one_to_three <int, nothing, nothing> y; // whatever 'nothing' needs to be.
sizeof(y) == 4;
T1 will never be 'nothing'
T2 will only be 'nothing' if T3 also is 'nothing' (if it works with T3 not being nothing, that's fine, but it won't get used that way).
Portability is a non-issue, this only needs to work in VS (2010 and higher). The 'real' solution will need up to T10, I have a solution working with SFINAE, but it takes very very long to compile and it's getting very unwieldy if you would need to add T11.
I know it's not an ideal type approach, but it is what it is, this is a necessity due to linking with a legacy API which we don't have control over.