C++ cast vector template

Question

I want to cast BaseClass to DerivedClass std vector using a template:

template<typename T, typename U>
vector<U> CastTo<U>(vector<T> &input)
{
    vector<U> output;
    for (auto obj : input)
    {
        output.push_back(dynamic_cast<U>(obj));
    }
    return output;
}

Is this possible?

Right now the template is not being recognized and I not able to use it. Is there a mistake in the proposed way?

The usage is:

vector<BaseClass*> vec;
vector<DerivedClass*> dVec = CastTo<DerivedClass*>(vec);

[mcve], please. Nothing stands out as overly wrong in your snippet. — StoryTeller - Unslander Monica, Mar 19 '18 at 10:22
What do you mean by "the template is not being recognized"? Do you get build errors? Then please show us the errors (copy-pasted, in full and complete). — Some programmer dude, Mar 19 '18 at 10:23

score 4 · Answer 1 · answered Mar 19 '18 at 10:24

4

template<typename T, typename U>
vector<U> CastTo<U>(vector<T> &input)
                ^^^

Remove that <U> (why the ALL_UPPERCASE that's for macros?). Also, be aware that your example call might be confusing T and U.

answered Mar 19 '18 at 10:24

Ulrich Eckhardt

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1

Good spot on that. Tricky brackets – StoryTeller - Unslander Monica Mar 19 '18 at 10:25
Okey, but how it is possible to specify the desired derived class? I naïvely wrote resembling what I am used to (C#) – Sturm Mar 19 '18 at 10:26
@Sturm `CastTo` still specifies the first template parameter. There is no extra `` in the definition, which is a different syntax to C# – Caleth Mar 19 '18 at 10:31
I'm aware of that, @Caleth, hence the last remark. Since the OP doesn't provide an MCVE, I'm not going to do all their work though. – Ulrich Eckhardt Mar 19 '18 at 10:34

The Vee · Accepted Answer · 2018-03-19T12:03:20.460

Two things,

don't use the <U> in the definition of CastTo<U>,
swap the order of T and U in the template line.

The first is just how the syntax for function templates goes in C++: no template brackets after their name. (But you will need one when calling the function.)

The second is more tricky. If you had a template<typename T, typename U> function and called CastTo<DerivedClass*>(vec), then DerivedClass* would be matched to the first parameter, T, leaving no way of determining U:

CastTo<DerivedClass*, ???>(vec): vector<DerivedClass*> &input -> ???
       ===== T =====  =U=                ↑ T

> template argument deduction/substitution failed:
> couldn't deduce template parameter ‘U’

If it's template<typename U, typename T>, then U will be DerivedClass* and T can be found from the function's argument vec:

1. CastTo<DerivedClass*, ???>(vec): vector<???> &input -> vector<DerivedClass*>
          ===== U =====  =T=                     ↑ vec            ↑ U
2. vec is a vector<BaseClass*>  =>  T = BaseClass*

> OK

This works (minimal example using primitive types and a C-style cast):

#include <vector>

using namespace std;

template<typename U, typename T>
vector<U> CastTo(vector<T> &input)
{
    vector<U> output;
    for (auto obj : input)
    {
        output.push_back(U(obj));
    }
    return output;
}

int main() {
  vector<float> v{2,3,5};
  vector<int> w = CastTo<int>(v);
  return w[2];  // returns 5
}

C++ cast vector template

2 Answers2