如何正确使用std :: function作为谓词

Question

I want to use std::function as a predicate in a templated helper function but I get an error:

 main.cpp(15): error C2672: 'Any': no matching overloaded function found main.cpp(15): error C2784: 'bool Any(const std::vector<T,std::allocator<_Ty>> &,const std::function<bool(const T &)>)': could not deduce template argument for 'const std::function<bool(const T &)>' from 'main::<lambda_1b47ff228a1af9c86629c77c82b319f9>' main.cpp(7): note: see declaration of 'Any' 

#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>

template <typename T>
bool Any(const std::vector<T>& list, const std::function<bool(const T&)> predicate)
{   
    return std::any_of(list.begin(), list.end(), [&](const T& t) { return predicate(t); });
}

int main()
{
    auto myList = std::vector<int>{ 1, 2, 3, 4, 5 };
    if (Any(myList, [](int i) { return i % 5 == 0; }))
    {
        std::cout << "Bingo!" << std::endl;
    }
    else
    {
        std::cout << "Not found :(" << std::endl;
    }

    return 0;
}

I don't understand what I'm doing wrong.

Answer 1

I don't understand what I'm doing wrong.

You're passing a lambda to Any , but the implicit conversion (from lambda to std::function ) won't be considered in template argument deduction , which makes T can't be deduced on the 2nd function argument.

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution , which happens later.

You can use std::type_identity (since C++20) to exclude the 2nd argument from deduction; then the template parameter T would be deduced on only the 1st argument, then the code works fine. eg

template <typename T>
bool Any(const std::vector<T>& list, const std::function<bool(const std::type_identity_t<T>&)> predicate)
//                                                                  ^^^^^^^^^^^^^^^^^^^^^^^
{   
    return std::any_of(list.begin(), list.end(), [&](const T& t) { return predicate(t); });
}

LIVE

If you can't apply C++20, you can refer to the linked page above which gives a possible implementation, it's not hard to implement your own type_identity . If you don't stick to std::function , you can just add another template parameter for predicate.

template <typename T, typename P>
bool Any(const std::vector<T>& list, P predicate)
{   
    return std::any_of(list.begin(), list.end(), predicate);
}

LIVE

Or make it more general to work with other containers or raw arrays.

template <typename C, typename P>
bool Any(const C& c, P predicate)
{   
    return std::any_of(std::begin(c), std::end(c), predicate);
}

LIVE

Answer 2

Correct Code:

#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>

template <typename T>
bool Any(const std::vector<T>& list, const std::function<bool(const T&)> predicate)
{   
    return std::any_of(list.begin(), list.end(), [&](const T& t) { return predicate(t); });
}

int main()
{
    auto myList = std::vector<int>{ 1, 2, 3, 4, 5 };
    if (Any<int>(myList, [](int i) -> bool { return i % 5 == 0; }))
    {
        std::cout << "Bingo!" << std::endl;
    }
    else
    {
        std::cout << "Not found :(" << std::endl;
    }

    return 0;
}

if (Any<int>(myList, [](int i) -> bool { return i % 5 == 0; }))

1. You need to specify template argument (Any)
2. You forgot to add return type of lambda