Template type deduction C++

Yeladia 0
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I am getting an error qualified-id in declaration before ‘<’ token from the following code:

// g++ -std=c++20 example.cpp
#include <iostream>

template <typename U = int>
struct Example {
    template <typename T>
    static void execute() {
        std::cout << "Hey" << std::endl;
    }
};

int main() {
    Example::execute<float>();
}

When I include the type for Example, such as Example<int>::execute<float>() it compiles successfully. Shouldn't the compiler be able to deduce the type since I specified it as default value?

c++
1 Answers

Class template argument deduction only applies when creating objects.

That is Example e; will deduce Example<int> e; via the default argument.

You are not creating an object though, and Example is not a class. You must include a template argument list. In this case, it can be empty though, since the template argument includes a default:

Example<>::execute<float>();
Miles Budnek0
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