显式指定泛型lambda的operator()模板参数是否合法？相关代码是否合规？

Great questions! Let's break this down clearly:

1. Is explicitly specifying template parameters for a generic lambda's operator() legal? #

This depends on which C++ standard you're targeting:

• C14/C17: This wasn't part of the official standard, but it was a widely adopted compiler extension supported by major tools like GCC and Clang. Many developers used it even before it was standardized.
• C++20 and newer: Yes, this is fully legal and part of the standard. The feature was formalized via a standard proposal (P0428R2), which explicitly allows calling a generic lambda's templated operator() with explicit template arguments.

2. Is the provided C++ code standard-compliant? #

Let's look at your code again:

#include <iostream>
int main() {
    [](auto v){ std::cout << v << std::endl; }.operator()<int>(42);
}

• When compiled with -std=c++14: This code isn't strictly standard-compliant in C14, but it works because GCC 7.2.0 and Clang 3.8.0 include this as an extension. The C14 standard didn't explicitly permit explicit template arguments for the generic lambda's operator(), but compilers added support because it was a useful feature.
• When compiled with -std=c++20 or later: This code is 100% standard-compliant. The feature is now part of the official language spec.

If you need strict C14/C17 compliance, you can achieve the same effect without explicit template arguments by letting deduction do its job:

// This is standard-compliant in C++14 and later
[](auto v){ std::cout << v << std::endl; }(42);

The compiler will automatically deduce v as int here, which gives you the exact same behavior as your explicit version.

内容的提问来源于stack exchange，提问作者Constructor