How to deduce template argument for lambda, in function overload?-CodePudding

How should I modify my present function signature

template<class TypeData,typename TypeFunc1 = Identity,typename TypeFunc2>
bool isPrime(const TypeData& n,TypeFunc1 calcSqrt = {},TypeFunc2 isDivisible = [](const TypeData& a,const TypeData& b) {return a%b==0;},const bool debug = false)

in order to be called from

auto fSqrt = [](decltype(n) v) {return std::sqrt(static_cast<float>(v));};
std::cout<<(isPrime(n,fSqrt)?"Positive":"Negative")<<'\n';

Visual Studio 2019 gives

C2783 'bool isPrime(const TypeData &,TypeFunc1,TypeFunc2,const bool)': could not deduce template argument for 'TypeFunc2'

Everything fine without TypeFunc2 isDivisible = [](const TypeData& a,const TypeData& b) {return a%b==0;}, though.

What is the correct syntax for passing default lambda?
Kindly help me, please.

CodePudding user response：

The problem is that default arguments don't contribute to the deduction of types in template parameters.

This can be showed with this primary example:

template <typename X>
void f(X a = 2);

int main() {
    f(); // The compiler spits out an error since it cannot determine the type 'X' holds
}

What you'd need to do in this scenario is to use function overloading instead:

// ...

template<class TypeData, typename TypeFunc1, typename TypeFunc2>
bool isPrime(const TypeData& n,
             TypeFunc1 calcSqrt,
             TypeFunc2 isDivisible,
             const bool debug);

template<class TypeData>
bool isPrime(const TypeData& n) {
    return isPrime(n, Identity{}, [](const TypeData& a, const TypeData& b) {
            return a % b == 0;
        }, false);
}

// ...

CodePudding user response：

You need to add a default type for your TypeFunc2, for example:

template<class TypeData,
         typename TypeFunc1 = Identity,
         typename TypeFunc2 = bool(*)(const TypeData&,const TypeData&)>
                            //^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
bool isPrime(const TypeData& n,
             TypeFunc1 calcSqrt = {},
             TypeFunc2 isDivisible = [](const TypeData& a,const TypeData& b) {return a%b==0;},
             const bool debug = false) {
  // ...
}

CodePudding user response：

Ruks's answer is actually very good. Another option is to use std::function, it might suit your needs as well:

template<class TypeData,typename TypeFunc1=Identity,typename TypeFunc2 = std::function<bool(const TypeData&, const TypeData&)>>
bool isPrime(
    const TypeData& n,
    TypeFunc1 calcSqrt = {},
    TypeFunc2 isDivisible = [](const TypeData& a,const TypeData& b) {return a%b==0;},
    const bool debug = false);