Declaration of Function before defining causing error in C

Question

I have been trying to write a code that can check if there is the sequence of 123 in an array or not for which I made a check function which does that work. But declaring this function before main function is causing problems with compiling when I am writting the arguments in it.

#include<conio.h>
#include<stdio.h>
#include<stdlib.h>

int check(int, int);      /* This line is cauing trouble */

void main()
{   int arr_size;
    int a[]={0,1,2,1,2,1,4,5,1,2,3,4,5};
    arr_size = sizeof(a)/sizeof(a[0]);
    printf("%d",check(a, arr_size));

}

int check(int a[], int arr_size)
{
    int i;
    for(i=0;i<arr_size-1; i++)
    {
    if(a[i]==1 && a[i+1]==2 && a[i+2]==3)
        {
            return 1;

        }
    }
    return 0;
}

The screenshot of the error is attached along.

Declaration part is not causing any problem and the code is working fine when I am not writting any argument in it as shown below.

int check();  

I expected that while declaring function it should take the prarameters which is not the case here. Guidance would be appreciated.

Answer 1

You have a conflict there.

int and int * (or, int []) are not the same types.

Update your forward declaration to be

int check(int *, int); 

That said, given the usage of arr_size, you need to change the loop condition from

for(i=0;i<arr_size-1; i++)

to

for(i=0;i<arr_size-2; i++)

since you're using [i+2] as one of the index inside the loop.

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Regarding the reason for int * (a pointer) to be equivalent to an array in this case, quoting C11, chapter §6.7.6.3

A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, [...]