C ++ 14和更高版本是否允许Lambda函数使用默认参数？ 如果可以，怎么办？

Question

Does lambda functions changes to accept default arguments in c++ 14?

This example doesn't work on C++ 11:

int main() {

    int i = 34;
    auto RectSurf = [i](int length = 0, int width = 0){ return length * width;};
    cout << RectSurf(10) << endl;


    std::cout << std::endl;
}

But it works fine on C++ 14 and above. So is it correct or not? Because C++ prime 5ed said that: "Passing Arguments to a Lambda As with an ordinary function call, the arguments in a call to a lambda are used to initialize the lambda's parameters. As usual, the argument and parameter types must match. Unlike ordinary functions, a lambda may not have default arguments (§ 6.5.1, p. 236). Therefore, a call to a lambda always has as many arguments as the lambda has parameters. Once the parameters are initialized, the function body executes.

As an example of a lambda that takes arguments, we can write a lambda that behaves like our isShorter function:" From C++ Prime 5ed.

But the same code on C++14 and above works fine! But Also I've read here in SO That lambda functions can have a default parameter on condition that no variable has been captured in Lambda's capture list.

Answer 1

Default Arguments in Lambda Expressions were explicitly added to the Standard with C++14, and were not permitted in C++11. You can see this if you run the code against two different compiler flags in GCC , one for C++11, one for C++14, and compile using the -Wpedantic flag, you'll see that when compiled for C++11, the compiler issues a warning (GCC still supports it, but tells you it's not standard-compliant) whereas in C++14, there's no warning.

So yes. In C++14 and later, Default Arguments in a Lambda Expression are valid.