CODEWITHSUNDEEP
cfileio
mh234
mh234

Reputation: 47

Reading line by line from a file in C

What I am trying to do is print out the contents of a file line by line. I run the program in terminal by doing: ./test testText.txt. When I do this, random characters are printed out but not what is in the file. The text file is located in the same folder as the makefile. What's wrong? 

#include <stdio.h>

FILE *fp;
int main(int argc, char *argv[])
{


char line[15];
fp = fopen(*argv, "r");

while((fgets(line, 15, fp)) != NULL)
    {
        printf(line);
        printf("\n");
    }


}

Upvotes: 0

Views: 331

Answers (4)

Mahonri Moriancumer
Mahonri Moriancumer

Reputation: 6003

Q: What's wrong?

A humble critique:

#include <stdio.h>

FILE *fp; // Perhaps this should be declared inside main?

int main(int argc, char *argv[])
   {
   char line[15]; // Are the file lines all 14 characters or less?  (seems small)

   fp = fopen(*argv, "r"); // Opening the binary executable file (argv[0])? Intereting.
   // Should check here to ensure that fopen() succeeded.      

   while((fgets(line, 15, fp)) != NULL)

OK... well, remember that this isn't a text file.. it's an executable (due to *argv). This will read some wacky (but not random) characters from the executable.

      {
      printf(line);  // Bad practice.  Should be: printf("%s", line);

Ok... now print the wacky characters? 

      printf("\n");  // Redundant. The '\n' characters will be supplied in 'line'.
      }

   // fclose() call missing.

   // Integer return value for main() is missing.
   }

Here is (perhaps) what was actually intended:

#include <stdio.h>
#include <errno.h>

int main(int argc, char *argv[])
   {
   int rCode = 0;
   FILE *fp = NULL;
   char line[255+1];

   if(argc != 2)
     {
     printf("Usage: %s {filepath}\n", *argv);
     goto CLEANUP;
     }

   errno=0;
   fp = fopen(argv[1], "r");
   if(NULL == fp)
      {
      rCode=errno;
      fprintf(stderr, "fopen() failed. errno:%d\n", rCode);
      goto CLEANUP;
      }

   while(fgets(line, sizeof(line), fp))  /* --As per 'chux' comment */
      printf("%s", line);

 CLEANUP:

   if(fp)
      fclose(fp);

   return(rCode);
   }

Or, if the intent is truly to print the content of the executable, perhaps this:

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

int main(int argc, char *argv[])
   {
   int rCode = 0;
   FILE *fp = NULL;
   off_t offset = 0;

   errno=0;
   fp = fopen(*argv, "r");
   if(NULL == fp)
      {
      rCode=errno;
      fprintf(stderr, "fopen() failed. errno:%d\n", rCode);
      goto CLEANUP;
      }

   for(;;)
     {
     char line[16];
     size_t bytesRead;
     int index;
     char ascii[16+1];

     memset(ascii, 0, sizeof(ascii));
     bytesRead = fread(line, 1, sizeof(line), fp);
     if(0==bytesRead)
        break;

     printf("   %08zX | ", offset);
     for(index=0; index < bytesRead; ++index)
        {
        printf("%02hhX%c", line[index], 7==index ? '-' : ' ');
        ascii[index] = isprint(line[index]) ? line[index] : '.';
        }

     printf("%*s   %s\n", (16 -index) * 3, "", ascii);
     offset += bytesRead;
     }

   if(errno)
      {
      rCode=errno;
      fprintf(stderr, "fgets() failed.  errno:%d\n", errno);
      }

CLEANUP:

   if(fp)
      fclose(fp);

   return(rCode);
   }

Upvotes: 1

Clifford
Clifford

Reputation: 93476

The first argument passed on the command line is at argv[1], while *argv refers to argv[0]. argv[0] contains the filename of the executable - you are printing out the content of the executable.

The following code prints out the entire argv[] array, then reads your file and prints it.

#include <stdio.h>

int main( int argc, char *argv[] )
{
    for( int i = 0; i < argc; i++ )
    {
        printf( "argv[%d] : %s\n", i, argv[i] ) ; 
    }

    if( argc >= 2 )
    {
        FILE* fp = fopen( argv[1], "r" ) ;
        if( fp != NULL )
        {
            char line[15];

            while( fgets( line, sizeof(line), fp ) != NULL )
            {
                printf( "%s", line ) ;
            }
        }
    }

    return 0 ;
}

Note that fgets() will read an entire line including the , so there is no need to print '\n', especially because with only 15 characters, your line buffer may well not contain an entire line. Note also the tighter localisation of variables - your code needlessly made fp global.

Other refinements are the safe use of the array size rather than literal 15, and the use of a literal constant string for the format specifier. You should avoid passing a variable string for the printf() format string - if your input itself contains format specifiers, printf() will try to read data from arguments that do not exist with undefined results.

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726579

When I do this, random characters are printed out but not what is in the file

These characters are not random, and in fact they are coming from a file. It's not the file that you are trying to read, though - it's the executable file which you are running.

*argv represents the name of the executable; add this line to see what's in *argv:

printf("%s\n", *argv);

The actual command line arguments start at argv[1], so you need

fp = fopen(argv[1], "r");

Upvotes: 1

eitank
eitank

Reputation: 330

your file name found at index 1 of argv.

if (argc <= 1) {
    printf("no file was given\n");
    exit(-1);
}

// open file from argv[1]
// ...

Upvotes: 0

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