I get an error saying, "exit status 1 expected initializer before 'void'." I will be wiring a switch to the Arduino that controls the voltage. When the I turn the switch it can adjust the voltage from 0 to 5. The overall task is to print the output voltage. The print is to be a multiplication table of the voltage.
I created this code:
void setup()

void loop(){
  
  int (n,i,j,result);
 
  printf(&quot;enter N Value&quot;);

  scanf (&quot;%d&quot;, &amp;n);

  for (i=1; i&lt;=n; i++);

  {
    for (j=1; j&lt;=n;j++);

    {
      result= (i*j);

      printf(&quot;%d&quot;, result);

    }
    printf(&quot;\n&quot;);

  }
  return 0;

}

But i get the error exit status 1
expected initializer before 'void' when I try to verify it. what does it mean. How do you fix it?
Functions must have a body. So instead of:
void setup()

You need:
void setup()
  {
  // do nothing
  }


You have other issues, for example this is not C++:
  int (n,i,j,result);

That should be:
  int n,i,j,result;

Also you are returning 0, when void functions don't return a value.
Also, here:
  printf(&quot;enter N Value&quot;);

  scanf (&quot;%d&quot;, &amp;n);

Where is this value coming from? There isn't a keyboard on the Arduino.
See: scanf

The C library function int scanf(const char *format, ...) reads formatted input from stdin.

The Arduino is a microprocessor. It doesn't have stdin.