is it possible to construct variadic arguments for function by overloading operator comma of the argument? i want to see an example how to do so.., maybe something like this:
template <typename T> class ArgList {
public:
ArgList(const T& a);
ArgList<T>& operator,(const T& a,const T& b);
}
//declaration
void myFunction(ArgList<int> list);
//in use:
myFunction(1,2,3,4);
//or maybe:
myFunction(ArgList<int>(1),2,3,4开发者_开发百科);
It is sort-of possible, but the usage won't look very nice. For exxample:
#include <vector>
#include <iostream>
#include <algorithm>
#include <iterator>
template <class T>
class list_of
{
std::vector<T> data;
public:
typedef typename std::vector<T>::const_iterator const_iterator;
const_iterator begin() const { return data.begin(); }
const_iterator end() const { return data.end(); }
list_of& operator, (const T& t) {
data.push_back(t);
return *this;
}
};
void print(const list_of<int>& args)
{
std::copy(args.begin(), args.end(), std::ostream_iterator<int>(std::cout, " "));
}
int main()
{
print( (list_of<int>(), 1, 2, 3, 4, 5) );
}
This shortcoming will be fixed in C++0x where you can do:
void print(const std::initializer_list<int>& args)
{
std::copy(args.begin(), args.end(), std::ostream_iterator<int>(std::cout, " "));
}
int main()
{
print( {1, 2, 3, 4, 5} );
}
or even with mixed types:
template <class T>
void print(const T& t)
{
std::cout << t;
}
template <class Arg1, class ...ArgN>
void print(const Arg1& a1, const ArgN& ...an)
{
std::cout << a1 << ' ';
print(an...);
}
int main()
{
print( 1, 2.4, 'u', "hello world" );
}
Operators have a fixed number of parameters. You cannot change that. The comma operator takes two arguments. So no. You can roll a custom, cascading version though, with some effort.
Maybe something like this:
class MyArgList {
public:
typedef std::list<boost::any> ManyList;
template <typename T>
MyArgList& operator, (const T& val) {
elems.push_back(val);
return *this;
}
ManyList::iterator begin() {return elems.begin();}
...
private:
ManyList elems;
};
Usage would be:
void foo(MyArgList& list);
foo((myArgList(),1,2,3,4,5));
No, it isn't. The list of values separated by the comma operator will be evaluated as a single value. For example:
1,2,3
will result in a single value, 3.
精彩评论