C结构函数语法-传递元素，返回结构

Question

I'm trying to create a function that accepts an array and two integers to manipulate and return in a struct.

What i have looks like this:

#include <stdio.h>

struct Results {
    int *A; // Pointer para o Array
    int N; // Comprimento do Array
};

int k, n;
struct Results solution(int A[], int N, int K);

int main(void){
    int a[] = {1,2,3};
    struct Results out;

    k = 1;
    n = sizeof(a)/sizeof(a[0]);
    printf("n = %d \n", n);

    out = solution(int a[], int n, int k);
    // EXPECTED EXPRESSION !!
}

struct Results solution(int A[], int N, int K) {
    struct Results outp;
    outp.A = A;
    outp.N = N;
    return outp;
};

I can't pass from this point, the compiler tells me that a expression is expected when I declare the function.

I think this might be a basic syntax error...

Answer 1

You are mixing the syntax for defining or declaring a function with the actual call to that function:

out = solution(a, n, k);

The types of the parameters are only present in the prototype, not in the call.