Solved

compile_error

Posted on 2004-10-05
2
222 Views
Last Modified: 2010-04-15
I have this compile error, not sure what it means "error : two or more data types in declaration of 'main' "

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

/* structure definitions */
    struct process
    {
       char proc[2];
       int D;       /* disk drives */
       int T;       /* tape drives */
       int P;       /* printer */
       int G;       /* graphical platters */
       int Dm;    /* disk drives maximum */
       int Tm;    /* tape drives maximum */
       int Pm;    /* printer maximum */
       int Gm;    /* graphical platters maximum */
    }
    /* the main client program */
    int main ()
   {
   /*
     Total Resources:
     D --- 13 Disk drives
     T --- 15 Tape drives
     P --- 12 Printer
     G --- 13 Graphical platters
    */
     
      int D = 18;
      int T = 17;
      int P = 11;
      int G = 13;
      int i = 0, j = 0, t = 0;
      int Allocated[10][4], Maximum[10][4], Need[10][4], Work[4];
      int finish[10];
      int Finish_Stack[10];
      struct process proc;
   
      FILE *pt;   /* File pointer */
      char cmd;      /* detect  user command for continue or quit */
   
      if(pt = fopen("proc.dat", "r")) == NULL)
      {
         printf("error: fopen() failed on data.txt\n");
         exit(1);    /*file open error handler */
      }
      printf("\nAllocation\tMaximum \tNeed\n");
   
      while(fscanf(pt, "%s%d%d%d%d%d%d%d%d", proc.proc, &proc.D, &proc.T,&proc.P,
                    &proc.G, &proc.Dm, &proc.Tm, &proc.Pm, &proc.Gm)!=EOF)
      {
         Allocated[i][0] = proc.D;
         Allocated[i][1] = proc.T;
         Allocated[i][2] = proc.P;
         Allocated[i][3] = proc.G;
         Maximum[i][0] = proc.Dm;
         Maximum[i][1] = proc.Tm;
         Maximum[i][2] = proc.Pm;
         Maximum[i][3] = proc.Gm;
         
        /* Calculate the Need : need[i,j]=max[i,j]-allocation[i,j] */
         Need[i][0] = Maximum[i][0] - Allocated[i][0];
         Need[i][1] = Maximum[i][1] - Allocated[i][1];
         Need[i][2] = Maximum[i][2] - Allocated[i][2];
         Need[i][3] = Maximum[i][3] - Allocated[i][3];
         D = D - Allocated[i][0];
         T = T - Allocated[i][1];
         P = P - Allocated[i][2];
         G = G - Allocated[i][3];
         
        /* set finish[i] = false for i = 1...n */
         finish[i] = 0;
         printf("\n( %d %d %d %d)\t( %d %d %d %d )\t( %d %d %d %d )\n", Allocated[i][0],
            Allocated[i][1], Allocated[i][2], Allocated[i][3], Maximum[i][0],
            Maximum[i][1], Maximum[i][2], Maximum[i][3], Need[i][0],
            Need[i][1], Need[i][2], Need[i][3]);
         i++;
      } /* end while */
   
    /* set work = available */
      printf("\nAvailable: \n");
      Work[0] = D;
      Work[1] = T;
      Work[2] = P;
      Work[3] = G;
      printf("\n%d %d %d %d\n", Work[0], Work[1], Work[2], Work[3]);
    /*-------------------------------------------------------------*/
   
      printf("\n==================================================================\n");
      i = 0;
      j = 0;
      printf("\n\npress <ENTER> to the next safe statemet\t");
      cmd = getchar();
   
      while (cmd == '\n')
      {
         if((finish[0] == 1) && (finish[1] == 1) && (finish[2] == 1) &&
         (finish[3] == 1) && (finish[4] == 1) &&(finish[5] == 1) && (finish[6] == 1) && (finish[7] == 1) && (finish[8] == 1) && (finish[9] == 1))
         {
            printf("\n\t=================================================\n");
            printf("\n\t\tAll Processes completed successfully\n");
            printf("\t\t\tWith Such a Sequence : \n\n\t\t( ");
            for(t = 0; t < 10; t++)
            {
               printf("P%d ", Finish_Stack[t]);
            }
            printf(")\n\n");
            printf("\t\t\tThe System is SAFE !\n");
            printf("\n\t=================================================\n");
            printf("\n\nProgram Terminated ... \n\n\n");
            exit(1);
         } /* if */
         for(i = 0; i < 10; i++)
         {
            printf("\nChecking Process %d", i);
            printf("\nWork ( %d %d %d %d )\n", Work[0], Work[1], Work[2], Work[3], i);
            printf("Need ( %d %d %d %d )\n", Need[i][0], Need[i][1], Need[i][2], Need[i][3]);
            if((finish[i] == 0) && (Need[i][0] <= Work[0]) && (Need[i][1] <= Work[1])
                                && (Need[i][2] <= Work[2]) && (Need[i][3] <= Work[3]))
            {
               printf("Need <= Work :\n");
               printf("So Doing Process %d\n", i);
               printf("..\n");
               printf("....\n");
               printf("..........\n");
               printf("......................\n");
           
               Work[0] = Work[0] + Allocated[i][0];
               Work[1] = Work[1] + Allocated[i][1];
               Work[2] = Work[2] + Allocated[i][2];
               Work[3] = Work[3] + Allocated[i][3];
               finish[i] = 1;
           
               printf("NOW Process %d DONE !\n\n", i);
               Finish_Stack[j] = i;
               printf("The Process have been done : ");
               for(t = 0; t < = j; t++)
               {
                  printf("P%d ", Finish_Stack[t]);
               } /* end inner for */
               
               printf("\n");
               j++;
               break;
            }
            else
            {
               if(finish[i] == 1)
               {
                  printf("This Work already been done\n");
                  printf("Jump to next process\n");
               }
               else
               {
                  printf("BECAUSE : There were Existing Need > Work\n");
                  printf("---> So, jump to next process\n");
               }
               if(i == 9)
               {
                  printf("\n\t===============================================================\n");
                  printf("\n\t\tAll Process have been checked\n");
                  printf("\t\tThere is no appropriate Process can be processing\n");
                  printf("\t\tSo There was NO Safe State so far\n");
                  printf("\n\t===============================================================\n");
                  printf("\n\nProgram Terminated ... \n\n\n");
                  exit(1);
               }
            }
         }/* end for */
         printf("\n\n\npress <ENTER> to the next safe statemet\t");
         cmd = getchar();
      } /* end while */
     
      /* close the file */
      fclose(pt);
     
      return(0);
   }



0
Comment
Question by:gbilios
2 Comments
 
LVL 45

Accepted Solution

by:
Kdo earned 100 total points
ID: 12226392

Pretty subtle....  :)

The message means that a declaration has more than one data type.  A simple one would be:

int float SomeData;

Obviously there is a conflict here that the compiler won't be able to resolve.


In your example, the compiler tells us where the problem lies -- the typing of the main function.  If you look closely, you'll note that the token immediately preceding "main" is "int".  So far so good.

But if you back up one more token you'll see a struct{} definition.  The compiler is confused because it thinks that you're trying to apply both types (struct and int) to main().


Now the good news.  Put a ';' after the struct definition and all of this should go away.  :)



Good Luck,
Kent

0
 

Author Comment

by:gbilios
ID: 12226645
i just realised that now.  I compiled it with gcc -ansi -Wall -o Safe Safe.c (SunC/C++ compiler) and okay.  

Thanks

0

Featured Post

Maximize Your Threat Intelligence Reporting

Reporting is one of the most important and least talked about aspects of a world-class threat intelligence program. Here’s how to do it right.

Join & Write a Comment

Suggested Solutions

Have you thought about creating an iPhone application (app), but didn't even know where to get started? Here's how: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Important pre-programming comments: I’ve never tri…
This is a short and sweet, but (hopefully) to the point article. There seems to be some fundamental misunderstanding about the function prototype for the "main" function in C and C++, more specifically what type this function should return. I see so…
Video by: Grant
The goal of this video is to provide viewers with basic examples to understand and use for-loops in the C programming language.
The goal of this video is to provide viewers with basic examples to understand and use switch statements in the C programming language.

760 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

20 Experts available now in Live!

Get 1:1 Help Now