user defined array function, I do not see the mistake

Question

The issue has been solved thanks to Eugene, however, I am still curious about what I did wrong with the user-defined functionI am a beginner and I do not see my mistake in this code, I am trying to print the array elements then swap the first and last element then print the values again, is it possible to swap without a user-defined function? if so how? thank you in advance. this is my code.

#include<stdio.h>
int swap(int);
int main()
{
    int size;
    int a[5]={10,20,30,40,50};
    for(size=0;size<5;size++)
    {
        printf("%d",a[size]);
    }
    swap(a);
    size=0;
    for(size=0;size<5;size++)
    {
        printf("%d",a[size]);
    }
}
int swap(int)
{
    int temp=a[0];
    a[0]=a[4];
    a[4]=temp;
}

Answer 1

The error message actually exactly tells you what is wrong:

Parameter name missing. a undefined... (Please do not post text error message as pictures. We cannot copy&paste it here form your picture.)

In function definition you must provide a name for each parameter. Otherwise you cannot use it inside that function:

int swap(int)
{
int temp=a[0];
a[0]=a[4];
a[4]=temp;
}

Here don't tell your compiler how that int parameter is called. And on the other side you don't tell the compiler what you think, a should be. It is not visible in scope.

To fix that, change like this:

int swap(int a[])
{
int temp=a[0];
a[0]=a[4];
a[4]=temp;
}

That will solve the error messages you showed us. But you should also get at least a warning about not retuning anything from a non-void function. You defined return type as int but don't return any int value. For a swap function there is normally no need to return anything.

Therefore use this:

void swap(int a[])
{
  int temp=a[0];
  a[0]=a[4];
  a[4]=temp;
}

Any decent beginners text book should clearly cover how function definitions have to look like. You should not skip the first chapters when learning a new language.