hn.ft.pris@gmail.com wrote:
Quote:
I've got following code test C++'s functor. For the sake of
easy-reading, I omit some declearations.
>
>
>
#include <algorithm>
#include <functional>
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using namespace std;
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template <typename Tclass Sum{
public:
Sum(T i=0):sum(i){};
inline void operator () (T x){
sum += x;
}
inline T output() const{
return sum;
}
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private:
T sum;
};
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int main( void ){
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vector<intvec(10, 1);
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Sum<intsum;
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sum = for_each(vec.begin(), vec.end(), Sum<int>()); .......(1)
sum = for_each(vec.begin(), vec.end(), sum); .........(2)
sum = for_each(vec.begin(), vec.end(), sum.operator()(int) ); ...(3)
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cout << sum.output() << endl;
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return 1;
}
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It's easy to understand that (2) works, because sum.operator()(int) is
called implicitly. (1) also works, it confuses me. Does Sum<int>()
create an implicit class Sum object?
There is nothing implicit about the function object. Although: it is a
temporary.
Quote:
On the other hand, Sum<int doesn't work.
Right: Sum<intjust denotes a type. It is not an expression that evaluates
to an object of that type.
Quote:
(3) fails, does it means the third argument of "for_each" couldn't be a
function pointer?
No, it just means you got the syntax wrong.
Quote:
What will the code be if I want to pass a function pointer here?
Well, just pass a function pointer. Something like:
void inc_5 ( int & a ) { a+=5; }
....
for_each( vec.begin(), vec.end(), &inc_5 );
(warning, I just typed it into the newsreader, so there may be typos in the
code.)
Best
Kai-Uwe Bux
