Why this fails?

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:

Quote:

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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>'

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Code:

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#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;
}

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plus2 is unspecialized - you must pass a specialized function.

Code:

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forAll(vec, plus2<T>);

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Replace T with type.

Thank you Elysia :)

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.

Quote:

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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.

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

--
Mats

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.

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

Quote:

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Originally Posted by MSDN

Deletes the element at the end of the vector.

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It isn't supposed to return a value.

Quote:

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Originally Posted by manav

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

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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

Quote:

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plus2 is unspecialized - you must pass a specialized function.

Code:

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forAll(vec, plus2<T>);

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Really? Does the compiler not deduce the template parameter?

Quote:

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Does the compiler not deduce the template parameter?

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

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

Quote:

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

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I am not sure how you came to that conclusion.

Quote:

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Originally Posted by laserlight

I am not sure how you came to that conclusion.

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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:

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

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I doubt the utility of it though.