How can I pass the return of a lambda function to another function in C++?

Question

I have an external function defined as extern "C" int64_t ASMFunction(int64_t sort);

Then I have a C++ function defined like this:

template <typename T>
void results(T&& sort) {
    ASMFunction([&](int64_t sort[60]) {for (int i = 0; i < sizeof(sort); i++) {
        sort[i] = rand() % 9223372036854775807;
    };
    return sort; }
    );
}

The reason I define it under a template is because otherwise I get the compiler error no suitable conversion function from "lambda []int64_t *(int64_t *sort)->int64_t *" to "int64_t" exists. Otherwise, I wouldn't even have the separate function to call my external ASMFunction, but would just do it in main(). With that, I get the compiler error Error C2664 'int64_t ASMFunction(int64_t)': cannot convert argument 1 from 'main::<lambda_2b1e6f6fde7359a6cb4cb68639dfae02>' to 'int64_t'. How can I use the C++ Lambda function to generate an array of random integers as a parameter for another function?

Please provide a [mcve]. Does this answer your question? [How to immediately invoke a C++ lambda?](https://stackoverflow.com/questions/44868369/how-to-immediately-invoke-a-c-lambda) — Thomas Sablik, Jul 16 '20 at 20:49
Like any other function "array" arguments, lambda "array" arguments are really pointers. Which means `sizeof(sort)` is misleading. — Some programmer dude, Jul 16 '20 at 21:14
@ThomasSablik that fixed it, but I'm still wondering if there's a way to include the lambda in main without having to use a template definition (because creating a whole new function makes it so much messier!) :( https://i.imgur.com/laPNBbi.png — eyeseaevan, Jul 16 '20 at 21:16
@cdhowie Nothing yet, it just stores it in the rcx register (first argument register for C++ function parameters), for my function in x64 Assembly. I was planning to write sorting algorithms in ASM as a learning exercise, but want to figure out how to use lambdas in C++ first. — eyeseaevan, Jul 16 '20 at 21:18

Manuel · Accepted Answer · 2020-07-16T22:01:06.790

You are trying to generate a list of numbers, but your ASMFunction only gets one.

If you want a list you can do this:

int64_t ASMFunction(int64_t * sort)
{
    std::cout << "Num: " << sort[1] << endl;
    return 0;
}

template <typename T, const int64_t N>
void results(T (&sort)[N]) {
    ASMFunction([&]() {for (int64_t i = 0; i < N; i++) {
                sort[i] = rand() % 9223372036854775807;
            };
            return sort;
                }()  // <- note you are calling the function to generate the list
        );
}

int main() {
    int64_t i[10];
    
    results(i);
    
    return 0;
}

If you want only one number:

int64_t ASMFunction(int64_t sort)
{
    std::cout << "Num: " << sort << endl;
    return 0;
}

template <typename T, const int64_t N>
void results(T (&sort)[N]) {
    ASMFunction([&]() { return rand() % 9223372036854775807; }() );
}

int main() {
    int64_t i[10];
    
    results(i);
    
    return 0;
}

The typename T is the type, in this case int64_t and the const int64_t N is to be able to have the array size deduced by the template. There will be a new template instance for each array size.

Thanks so much, exactly what I wanted to know. :). If you could also please explain what declaring the new typename T does, and why it's needed for this function, I would really appreciate it. — eyeseaevan, Jul 16 '20 at 21:29

How can I pass the return of a lambda function to another function in C++?

1 Answers1