Syntax Adjustment in C code block

I'm supposed to program an LCD screen in MC9S12E128 that outputs the a couple of names on it. The name changes each time the push buttons are pushed. This needs to be done with C language. I wrote this code so far but it seems to be wrong..

PHP Code:

---

#include <hidef.h> /* common defines and macros */
#include <mc9s12e128.h> /* derivative information */
#include < stdio.h>
#pragma LINK_INFO DERIVATIVE "SampleS12"

void initPort(void);
int main () {
PORTB = 0x00;
PORTA = 0x00;
PORTE = 0X00;
DDRA = 0XFF;
DDRE = 0xFF;
}

void main (){
char name1[8],name2[8],name3[8],name4[8];
int input; 

EnableInterrupts;
for(;;) { 

PORTE=0x01 ; 

DDRE = DDRA ;
}
char *name1 [8] = "Name1"; 
char *name2 [8] = "Name2" ;
char *name3 [8] = "Name3";
char *name4 [8] = "Name4" ;

switch ( input ) {
case 1: /* Note the colon, not a semicolon */
PORTA: name1 

break;
case 2: 

PORTA : name2 

break;
case 3:
PORTA ; name3 

break;

default: 
PORTA : name4
break;
}
getchar();
}
}
}
} 


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The debugger tells me that Return is Expected n stuff, any idea what could be wrong with my code? Would really appreciate your help..
Thanks!

Code:

---

int main () {
PORTB = 0x00;
PORTA = 0x00;
PORTE = 0X00;
DDRA = 0XFF;
DDRE = 0xFF;
}

---

Don't know as far as embedded devices and what have you go. But this is still C, you must return some integer. Usually returning 0 from main() means "no error".
You also can't have another function of the same name, there is no overloading/overriding in vanilla C. You have 2 main()s pick one. Note that void main() is "wrong" -- see the FAQ.

Just remember to be careful, with some compilers using void main() can cause the computer to physically catch fire.

Quote:

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Just remember to be careful, with some compilers using void main() can cause the computer to physically catch fire.

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Are you serious? XD
This is my first time working with C really, so im as clueless as hell. Got to fix the return problem, thanks a million for that Zac. Also change void main() to void body().. One curly bracket error to be fixed and it should all work. Here is how the code looks now..

PHP Code:

---

#include <hidef.h>      /* common defines and macros */
#include <mc9s12e128.h>     /* derivative information */
#include < stdio.h>
#pragma LINK_INFO DERIVATIVE "SampleS12"


void initPort(void);
int main () {
  
  
 
  PORTB = 0x00;
  PORTA = 0x00;
  PORTE = 0X00;
  DDRA = 0XFF;
  DDRE = 0xFF;
  return 0;
}

        void body (){
        char name1[8],name2[8],name3[8],name4[8];
        int input; 
                
  EnableInterrupts;
  
   for(;;) { 
            
     PORTE=0x01 ; 
     DDRE = DDRA ;
  }
  char *name1 [8] = "Name1"; 
  char *name2 [8] = "Name2"  ;
  char *name3 [8] = "Name3";
  char *name4 [8] = "name4" ;
 
   switch ( input ) {
        case 1:            /* Note the colon, not a semicolon */
           PORTA: name1 
      
            break;
        case 2:          
                  
         PORTA : name2 
            
            break;
        case 3:
       PORTA ; name3 
         
            break;
       
        default:            
            PORTA : name4
            break;
    }
    getchar();
}
}
}
} 


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Okay, this has been getting on my nerves lol.. It keeps on saying ' } ' missing, one the line marked below..

PHP Code:

---

#include <hidef.h>      /* common defines and macros */
#include <mc9s12e128.h>     /* derivative information */
#include < stdio.h>
#pragma LINK_INFO DERIVATIVE "SampleS12"


void initPort(void);
int main () {
  
  
 
  PORTB = 0x00;
  PORTA = 0x00;
  PORTE = 0X00;
  DDRA = 0XFF;
  DDRE = 0xFF;
  return 0;
}

        void body (){
        char name1[8],name2[8],name3[8],name4[8];
        int input; 
                
  EnableInterrupts;
  
   for(;;) { 
            
     PORTE=0x01 ; 
     DDRE = DDRA ;
  }
  char *name1 [8] = "Name1";               // Error indication came here
  char *name2 [8] = "Name2"  ;
  char *name3 [8] = "Name3";
  char *name4 [8] = "Name4" ;
 
   switch ( input ) {
        case 1:           
           PORTA: name1 
      
            break;
        case 2:          
                  
         PORTA : name2 
            
            break;
        case 3:
       PORTA ; name3 
         
            break;
       
        default:            
            PORTA : name4
            break;
    }
    getchar();
}
}
}
} 


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Where is it missing?

Priobably because you are declaring variables after you have started with the code - the actual error message is a bit confusing, but I think it's simply a compiler gets confused sort of situation.

Also, your code never calls "body".

--
Mats

You might want to properly indent your code:

Code:

---

#include <hidef.h>      /* common defines and macros */
#include <mc9s12e128.h>    /* derivative information */
#include <stdio.h>
#pragma LINK_INFO DERIVATIVE "SampleS12"

void initPort(void);

int main() {
    PORTB = 0x00;
    PORTA = 0x00;
    PORTE = 0X00;
    DDRA = 0XFF;
    DDRE = 0xFF;
    return 0;
}

void body() {
    char name1[8],name2[8],name3[8],name4[8];
    int input;

    EnableInterrupts;

    for(;;) {
        PORTE=0x01 ;
        DDRE = DDRA ;
    }

    char *name1 [8] = "Name1";              // Error indication came here
    char *name2 [8] = "Name2"  ;
    char *name3 [8] = "Name3";
    char *name4 [8] = "Name4" ;

    switch ( input ) {
    case 1:
        PORTA : name1
        break;
    case 2:
        PORTA : name2
        break;
    case 3:
        PORTA ; name3
        break;
    default:
        PORTA : name4
        break;
    }
    getchar();
}
}
}
}

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At a glance, you have three extra closing braces, and in your switch you probably want to assign to PORTA instead of using it as a label. The current error you have might be because you declared some variables (name1, name2, name3 and name4) in the middle of a block, and that is not allowed prior to the 1999 edition of the C standard.

Oh, and you probably want to call body() from main().