Why this fails?

**~~manav~~** · 05-28-2008

I have this code, if I change the function plus2 to a non-templated version, then, the code compiles and runs fine. But as it is, templated definition is not compiling. I get this error from MSVC .Net 2003:

c:\my\src\cpp\test-bed\test-bed.cpp(362) : error C2896: 'void forAll(T &,F)' : cannot use function template 'void plus2(T &)' as a function argument
c:\my\src\cpp\test-bed\test-bed.cpp(353) : see declaration of 'plus2'
c:\my\src\cpp\test-bed\test-bed.cpp(362) : error C2784: 'void forAll(T &,F)' : could not deduce template argument for 'T1 &' from 'std::vector<_Ty>'

Code:

#include <iostream>
#include <vector>

template <typename T, typename F>
void forAll(T& array, F f) {
    for (int i=0; i<array.size(); i++) {
        f(array[i]);
    }
}

template <typename T>
void plus2(T& v) {
    v += 2;
}

int main() {
    std::vector<int> vec;
    vec.push_back(10);
    vec.push_back(20);
    vec.push_back(30);
    forAll(vec, plus2);
    std::cout << vec[0] << " " << vec[1] << " " << vec[2] << std::endl;
    return 0;
}

**Elysia** · 05-29-2008

plus2 is unspecialized - you must pass a specialized function.

Code:

forAll(vec, plus2<T>);

Replace T with type.

**~~manav~~** · 05-29-2008

Thank you Elysia

**~~manav~~** · 05-29-2008

A very strange behavior of this program, when I changed it (shown in RED):

Code:

#include <iostream>
#include <vector>

template <typename T, typename F>
void forAll(T& array, F f) {
	for (int i=0; i<array.size(); i++) {
		f(array[i]);
	}
}

template <typename T>
void plus2(T& v) {
	v += 2;
}

int main() {
	std::vector<int> vec;
	vec.push_back(10);
	vec.push_back(20);
	vec.push_back(30);
	vec.pop_back();
	vec.pop_back();
	vec.pop_back();
	forAll(vec, plus2<int>);
	std::cout << vec[0] << " " << vec[1] << " " << vec[2] << std::endl;
	return 0;
}

Output:
10 20 30

And also, pop_back() does not return anything. Just decreases the size() of std::vector.

**matsp** · 05-29-2008

Originally Posted by manav

A very strange behavior of this program, when I changed it (shown in RED):

Code:

#include <iostream>
#include <vector>

template <typename T, typename F>
void forAll(T& array, F f) {
	for (int i=0; i<array.size(); i++) {
		f(array[i]);
	}
}

template <typename T>
void plus2(T& v) {
	v += 2;
}

int main() {
	std::vector<int> vec;
	vec.push_back(10);
	vec.push_back(20);
	vec.push_back(30);
	vec.pop_back();
	vec.pop_back();
	vec.pop_back();
	forAll(vec, plus2<int>);
	std::cout << vec[0] << " " << vec[1] << " " << vec[2] << std::endl;
	return 0;
}

Output:
10 20 30

And also, pop_back() does not return anything. Just decreases the size() of std::vector.

Accessing unused parts of a vector is undefined behaviour - anything could happen.

--
Mats

**Elysia** · 05-29-2008

Not strange - you simply so things that do not have a defined behavior, and therefore are expected to work in bad or unpredicted ways.
Try usingvec.at(0), etc instead.

**~~manav~~** · 05-29-2008

Strange is, that pop_back() does nothing. Except decrease the size(). Does not even return a value.

**Elysia** · 05-29-2008

Originally Posted by MSDN

Deletes the element at the end of the vector.

It isn't supposed to return a value.

**matsp** · 05-29-2008

Originally Posted by manav

Strange is, that pop_back() does nothing. Except decrease the size(). Does not even return a value.

A vector is not a stack where you need to get the data out via a pop operation. By the way, it may well have de-allocated the memory when you popped the last element - although it may not have done that, either - as an optimization, keeping the memory unless you actually do a specific resize operation, it will be better to keep the memory around, in case you start filling the vector again.

--
Mats

**MarkZWEERS** · 05-29-2008

plus2 is unspecialized - you must pass a specialized function.

Code:

forAll(vec, plus2<T>);

Really? Does the compiler not deduce the template parameter?

**laserlight** · 05-29-2008

Does the compiler not deduce the template parameter?

Not in this case, since no argument is supplied when passing the function pointer.

**MarkZWEERS** · 05-29-2008

Ok, so that's why a function object is useful?

**laserlight** · 05-29-2008

Ok, so that's why a function object is useful?

I am not sure how you came to that conclusion.

**brewbuck** · 05-29-2008

Originally Posted by laserlight

I am not sure how you came to that conclusion.

A function object could allow you to pass a generic functor without indicating the specialization -- the specialization would occur inside the functor call itself:

Code:

class plus2
{
public:
    template <typename T>
    T operator()(T &val)
    {
        return val  + 2;
    }
};

I doubt the utility of it though.

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