segmentation fault when opening a file

hi guys! pretty new here, just wanted to ask if someone can save me another 5hrs of research - what could cause segmentation fault while i open a file?
ps: it is in the same directory, and i tried entering whole path while starting program,.. nothing helps

thank you A LOT

here is the code:

Code:

---

int main(int argc, char *argv[])
{
  char *file_name = argv[1];
  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }
  FILE *config_file = fopen(argv[1], "r");
  if (config_file == NULL) {
    printf("Error: Cannot open file: %s\n", file_name);
    return 2;
  }
  printf("opened");

  fclose(config_file);
  return 0;
}

---

What does this code do?

Code:

---

if (argc != 2)
{
printf("Usage: ./a4 %s\n", file_name);
return1;
}

---

First you check if the argument counter is not equal to 2 and then use the second argument....

Quote:

---

Originally Posted by smilaf99

hi guys! pretty new here, just wanted to ask if someone can save me another 5hrs of research - what could cause segmentation fault while i open a file?
ps: it is in the same directory, and i tried entering whole path while starting program,.. nothing helps

thank you A LOT

---

First, always post your code as plain text without the colors.

Second, Compare my code below with your code:

Code:

---

#include <stdio.h>

int main(int argc, char *argv[])
{
  if (argc != 2)
  {
    printf("Usage: ./a4 file\n");

    return 1;
  }

  FILE *config_file = fopen(argv[1], "r");
  if (config_file == NULL)
  {
    printf("Error: Cannot open file: %s\n", argv[1]);
    return 2;
  }
  printf("opened\n");

  fclose(config_file);

  return 0;
}

---

Third, if you start the program without an input file, your first line in main() will or should assign NULL. Note the changes I also made.

Lastly, use indentation to make the code clear. Choose an indentation style and stick to it.

My version of your code works for me on Linux with gcc.

Quote:

---

Originally Posted by G4143

What does this code do?

Code:

---

if (argc != 2)
{
printf("Usage: ./a4 %s\n", file_name);
return1;
}

---

First you check if the argument counter is not equal to 2 and then use the second argument....

---

it is a snippet of code in which the error occurs.. and yes, i need exactly 2 arguments given, so im returning error value if it has more/less... i dont understand whats not clear :(

Quote:

---

Originally Posted by rstanley

First, always post your code as plain text without the colors.

Second, Compare my code below with your code:

Code:

---

#include <stdio.h>

int main(int argc, char *argv[])
{
  if (argc != 2)
  {
    printf("Usage: ./a4 file\n");

    return 1;
  }

  FILE *config_file = fopen(argv[1], "r");
  if (config_file == NULL)
  {
    printf("Error: Cannot open file: %s\n", argv[1]);
    return 2;
  }
  printf("opened\n");

  fclose(config_file);

  return 0;
}

---

Third, if you start the program without an input file, your first line in main() will or should assign NULL. Note the changes I also made.

Lastly, use indentation to make the code clear. Choose an indentation style and stick to it.

My version of your code works for me on Linux with gcc.

---

TBH i dont see anything changed.. just the indentation - which got deleted when i copy-pasted the code here.. so if it works for you, then the code is not the problem but the text file... thanks! ;)

Sorry... edited...

As rstanley explained you are using argv[1] BEFORE testing if it is valid:

Code:

---

  // Here! What happens if argc == 1?
  char *file_name = argv[1];

  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }
  ...

---

You should:

Code:

---

  char *file_name;

  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }

  file_name = argv[1];
  ...

---

Quote:

---

Originally Posted by flp1969

As rstanley explained you are using argv[1] BEFORE testing if it is valid:

Code:

---

  // Here! What happens if argc == 1?
  char *file_name = argv[1];

  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }
  ...

---

You should:

Code:

---

  char *file_name;

  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }

  file_name = argv[1];
  ...

---

---

Wrong!
In the second example, file_name is an uninitialized char pointer! What gets printed in the printf() if "argc != 2"??? Garbage!

See my code above for the preferred method!

Quote:

---

Originally Posted by smilaf99

TBH i dont see anything changed.. just the indentation - which got deleted when i copy-pasted the code here.. so if it works for you, then the code is not the problem but the text file... thanks! ;)

---

Again, don't attempt to access argv[1] BEFORE you test if you passed a filename!

Your code:

Code:

---

  char *file_name = argv[1];
  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }

---

My code:

Code:

---

  if (argc != 2)
  {
    printf("Usage: ./a4 file\n");
 
    return 1;
  }

---

You don't need "file_name" at all, just use argv[1] where appropriate. Compare my code to yours again.

Quote:

---

Originally Posted by rstanley

Wrong!
In the second example, file_name is an uninitialized char pointer! What gets printed in the printf() if "argc != 2"??? Garbage!

See my code above for the preferred method!

---

Ooops! My bad! You are right: The 3rd argument of fprintf, below, should be argv[0], as this:

Code:

---

if ( argc != 2 )
{
  fprintf( stderr, "Usage: %s <filename>\n", argv[0] );
  return EXIT_FAILURE;
}

---

This is a somewhat lame point, but you can't always trust argv[0] to not be NULL.

I've got two programs - 'b' that prints out if the first argv is NULL, and 'a' that calls it with no command line args.

Here's the code for 'b.c'

Code:

---

#include <stdio.h>


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


  printf("%i args\n",argc);
  if(argv[0] == NULL) {
      printf("argv[0] is NULL\n");
  }
  return 0;
}

---

Here's the code for 'a.c', that execs 'b' without ANY command line args:

Code:

---

#include <stdio.h>
#include <unistd.h>


char * const args[] = {
    NULL
};
char *env[] = {
    NULL
};


int main(int argc, char *argv[]) {
  execve("./b", args, env);
  fprintf(stderr, "Exec failed\n");
  return 0;
}

---

Quote:

---

Originally Posted by hamster_nz

This is a somewhat lame point, but you can't always trust argv[0] to not be NULL.
...

Code:

---

#include <stdio.h>
#include <unistd.h>

char * const args[] = {
    NULL
};
char *env[] = {
    NULL
};


int main(int argc, char *argv[]) {
  execve("./b", args, env);
  fprintf(stderr, "Exec failed\n");
  return 0;
}

---

---

Ahhhh... this is a very interesting point. Never thought about it this way, thanks hamster_nz!

Very interesting indeed! In my distro argv[0] is substituted by "". Using MSys2 or Cygwin exec??() fails, but it works on Linux. Of course, the only garantee is argv[argc]==NULL.

Quote:

---

Originally Posted by rstanley

Again, don't attempt to access argv[1] BEFORE you test if you passed a filename!

Your code:

Code:

---

  char *file_name = argv[1];
  if (argc != 2) {
    printf("Usage: ./a4 %s\n", file_name);
    return 1;
  }

---

My code:

Code:

---

  if (argc != 2)
  {
    printf("Usage: ./a4 file\n");
 
    return 1;
  }

---

You don't need "file_name" at all, just use argv[1] where appropriate. Compare my code to yours again.

---

yes, yes - i noticed a little while after replying, just because i focused more on the file opening part.. i usually test-case it with exactly 2 arguments, just like test cases our class provides.. so i kinda ignored it.. good point, i fixed it right after :)

indeed, i catch myself adjusting too much to my class' test cases (like i told in a reply before), than thinking outside of the box and going for the most efficient code overall - so not a lame point at all haha