Thread: Why do I have to fseek in this code?

There is a sentence in a text file and I am trying to insert a word into 9. character in the text file. For example,if the sentence is "Los Angeles is a good city,Paris is a good city" , my program will insert the text "CHPBL" into 9. character,after this modification the sentence in the file1.txt will look like:

"Los AngeleCHPBLs is a good city,Paris is a good city"


I did a little program to do it:

Code:

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


int main()
{
    


    char buffer[6];
    char buffer2[6];

    char buffer3[6];
    char word[6];
  
   



    strcpy( word, "CHPBL" );


    FILE* fp;
    if ((fp = fopen( "file1.txt", "r+t" )) == NULL)
    {
        printf( "File error!" );
        exit( EXIT_FAILURE );
    }


    fseek( fp, 9, SEEK_SET);     
    int counter=1;




    int offset3 = ftell(fp);
    fgets( buffer2, 6, fp);

    fseek( fp, offset3, SEEK_SET);
    
    fprintf( fp, "%s", word);
    int offset = ftell(fp);

    fgets( buffer, 6, fp);
   
     int counter2=0;

     while( !feof(fp) )
     {
         system( "pause" );
         
         int offset2 = ftell(fp);
       
         

         fseek( fp, offset2, SEEK_SET); //if I don't use this fseek                                                            
                                                    //program works wrong
                                                   //buffer3 takes wrong values 
         if ( fgets ( buffer3, 6, fp ) == NULL)
         {
             exit(0);
         }
         
         sayac++;
         printf( "%d. while loop", counter2);
         
         printf( "\nbuffer : %s \n ", buffer);
         printf(" \nbuffer3 : %s \n ",buffer3);
        
        


        if (counter==1)
        {
            
            fseek( fp, offset, SEEK_SET);
            
            fprintf( fp, "%s", buffer2);
            printf(" \nbuffer2 : %s \n", buffer2);
            
        }
        if (counter==0) fseek( fp, offset2, SEEK_SET);
        
        fprintf(fp, "%s", buffer);
        counter = 0;


        strcpy( buffer, buffer3);


}

This code works as I want but there is something which I am not able to understand.If I don't use fseek there ( I commented it in the code) it works wrong and I do not understand why I must use fseek there.

fseek adds an offset to the current possition of the file pointer. At a quick glance, that looks like offset2+offset2

fseek() - C Library Function Example

If you are curious, you can use printf to see what ftell returns; then add another one after the fseek to see where it is after.

Thanks for your answer

You said fseek adds an offset to the current position of the file pointer.If I use SEEK_CUR ( fseek(fp,offset2,SEEK_CUR)) you are right,but if I use SEEK_SET instead of SEEK_CUR, shouldn't it add offset from the position of beginning of the file?

I tried what you said:

Code:

         int offset2=ftell(fp);
         printf("\noffset2:%d\n",offset2);
         
        
         fseek(fp,offset2,SEEK_SET);


         int offset12=ftell(fp);
         printf("\noffset2 : %d\n", offset12 );

I used this code inside the code I wrote in my question,and both offset2 and offset12 printed the same values?

You are correct with the SEEK_SET [sorry]

If you had a file with the numbers 0 to 1000, what do you get when you read from the file?

Where is the program running, by the way? (i.e. Windows 8.1, Linux, PIC micro, ...)

Also, you should not be using feof in a control loop
FAQ > Why it's bad to use feof() to control a loop - Cprogramming.com

Thanks for your answer.

Where is the program running, by the way? (i.e. Windows 8.1, Linux, PIC micro, ...)

Program is running on Windows 7.

You are right about using feof() in loops,in my program I think it doesn't work actually, the lines:

Code:

if ( fgets ( buffer3, 6, fp ) == NULL)
         {
             exit(0);
         }

these lines do the job of checking if the end of file is reached in my code I think,feof which I used in my code doesn't work(or at least doesn't work right).