Why Does Type Deduction Fail with Lambda Functions in C Function Templates?

Type Deduction Pitfalls with Lambda Functions

In C , type deduction is a powerful feature that allows the compiler to infer the types of variables and expressions. However, it can encounter challenges when it comes to dealing with lambda functions and std::function objects.

Consider the following function template:

template<class A>
set<A> filter(const set<A>&amp; input, function<bool(A)> compare) {
    // Implementation omitted
}

When calling this function with a lambda function directly, such as:

filter(mySet, [](int i) { return i % 2 == 0; });

you may encounter an error stating that there is no matching function for the call. This occurs because type deduction cannot handle the lambda function as a direct argument to std::function.

The reason for this is that lambda functions are not considered functions in the strict sense, but rather function objects with a specific set of characteristics. The standard allows for lambdas to be converted to std::function objects with explicit parameter types and, in certain cases, function pointers. However, this does not elevate them to the same level as std::function.

To circumvent this limitation, there are several approaches you can take:

1. Convert the Lambda to a Function Object:

std::function<bool(int)> func = [](int i) { return i % 2 == 0; };
set<int> myNewSet = filter(mySet, func);

2. Provide the Template Type Argument Explicitly:

set<int> myNewSet = filter<int>(mySet, [](int i) { return i % 2 == 0; });

3. Use a Template Class:

template<class A, class CompareFunction>
set<A> filter(const set<A>&amp; input, CompareFunction compare) {
    // Implementation omitted
}

set<int> result = filter(myIntSet, [](int i) { i % 2 == 0; });

4. Utilize Dectype and Function Type Syntax:

template<class Value, class CompareType, class IndexType>
auto filter(const set<Value>&amp; input, CompareType compare, IndexType index) -> map<decltype(index(*(input.begin()))), Value> {
    // Implementation omitted
}

map<string, int> s = filter(myIntSet, [](int i) { return i % 2 == 0; }, [](int i) { return toString(i); });

By employing these strategies, you can successfully utilize lambda functions and std::function objects while taking into account the type deduction limitations of C .

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