CODEWITHSUNDEEP
cbus-error
Jack Kobil
Jack Kobil

Reputation: 13

Bus error in simple C program

I have the following simple program. And I am running "program file.txt" from the command line and getting "Bus error" with no output. I don't know what I'm doing wrong here.

#include<stdio.h>

int main(char *arg[])
{

    FILE *file = fopen(arg[0], "r");

    if (file == NULL) {
        printf("Cannot open file.\n");
        exit(1);
    }

    return 1;
}

Upvotes: 1

Views: 1343

Answers (4)

Jeeva
Jeeva

Reputation: 4663

argv[0] contains the application name. Try using argv[1] instead.

Upvotes: 0

Musa
Musa

Reputation: 97707

The prototype for the c entry function is

int main(int argc, char *arg[]);

So with your prototype you are actually trying to dereference an int to passing to fopen
Try this instead

#include<stdio.h>
int main(int argc, char *argv[])
{
    FILE *file = fopen(argv[1], "r");
    if (file == NULL) {
        printf("Cannot open file.\n");
        exit(1);
    }
    return 1;
}

Upvotes: 3

steveha
steveha

Reputation: 76725

The standard prototype for main() should look like this:

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

You declared your main() with just one argument, argv. But the system is passing a count of arguments as the first argument.

When you specify a single argument (the file file.txt), argc is set to 1. But your program is trying to use the integer 1 as a char **. That gives the bus error.

Here is an edited version of your program that works:

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

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

    FILE *file;

    if (argc != 2)
    {
        fprintf(stderr, "Usage: myprogram <filename>\n");
        exit(1);
    }

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

    if (file == NULL) {
        fprintf(stderr, "Cannot open file.\n");
        exit(2);
    }

    exit(0); // status 0 to signal no error
}

Changes:

  • I added #include <stdlib.h> to get a declaration of exit().

  • I check the number of arguments and print a Usage: string if it's not right.

  • I made the error messages print to stderr instead of standard output.

  • I changed return 1 to exit(0) to make it clear that the program is exiting successfully.

Upvotes: 1

RolandXu
RolandXu

Reputation: 3678

could you print out the value of arg[0], suppose the type of main is 

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

and the argv[0] is the name of process, the argv[1] is first argument.

Upvotes: 2

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