[BUG] parameter passing fail if out, forward, or move used as variable name

[filipsajdak]

In the current implementation of cppfront the following code:

main: () -> int = {
    out := 1;
    forward := 2;
    move := 3;
    inout := 4;
    copy := 5;
    in := 6;

    std::cout << out << std::endl;
    std::cout << forward << std::endl;
    std::cout << move << std::endl;
    std::cout << inout << std::endl;
    std::cout << copy << std::endl;
    std::cout << in << std::endl;
} 

Will compile and print:

1
2
3
4
5
6

When you try to pass these variables to the function:

main: () -> int = {
    out := 1;
    forward := 2;
    move := 3;
    inout := 4;
    copy := 5;
    in := 6;

    fun(out);
    fun(forward);
    fun(move);
    fun(inout);
    fun(copy);
    fun(in);
}

fun: (i : int) = {
    std::cout << i << std::endl;
}

it ends with a compilation error:

../tests/variable_names_passing_style.cpp2:9:5: error: no matching function for call to 'fun'
    fun(   );
    ^~~
../tests/variable_names_passing_style.cpp2:17:6: note: candidate function not viable: requires single argument 'i', but no arguments were provided
auto fun(cpp2::in<int> i) -> void;
     ^
../tests/variable_names_passing_style.cpp2:10:5: error: no matching function for call to 'fun'
    fun(       );
    ^~~
../tests/variable_names_passing_style.cpp2:17:6: note: candidate function not viable: requires single argument 'i', but no arguments were provided
auto fun(cpp2::in<int> i) -> void;
     ^
../tests/variable_names_passing_style.cpp2:11:5: error: no matching function for call to 'fun'
    fun(    );
    ^~~
../tests/variable_names_passing_style.cpp2:17:6: note: candidate function not viable: requires single argument 'i', but no arguments were provided
auto fun(cpp2::in<int> i) -> void;
     ^
3 errors generated.

The following cases failed:

    fun(out);
    fun(forward);
    fun(move);

Expectation

I would expect an error when using these keywords as variable names - I found this by accident trying to understand why I cannot pass variable out to the function:

out : *std::ostream = std::cout&;
// ...
print(out*, x, 0);

When additional keywords are added to the function argument list more variables will cause this issue (see: #230 (comment))

[hsutter]

Thanks! Yes, your examples should all work and now after the fix in the next commit. The production parameter-direction? expression does allow examples like fun(out) to not mean out with its special meaning.