运行C程序时在Linux中出现分段错误

Question

I am trying to simulate the cat command in Red Hat Linux . I am getting a segmentation fault when I run my program.

For example:

./a.out a > b

a contains hello. I expect hello to be copied in b .

My code is as follows:

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <fcntl.h>
#include <string.h>

int main(int argc,char *argv[])
{
    int f, fd, r;
    char buf[100];

    if (argc < 2)
    {
        printf("Error");
        return 0;
    }
    else
    {
        if (!strcmp(argv[2],">"))
        {
            f = open(argv[1],0,00777);

            if (f == -1)
                printf("no file");
            else
            {
                fd = creat(argv[3],00777);
                while( (r = read(f,buf,50)) > 0)
                    write(fd, buf, r);
            }
        }
    }
    return 0;
}

Why am I getting a segmentaion error?

I have a similar program where I open and create the file in the same manner, and that program is running, but this one is giving me a segmentation fault.

Answer 1

It's probably because the redirection is handled by the shell and not by your program, so argv[2] is NULL and argv[3] does not exist.

However you should use a debugger to find out what is really happening. And then add proper error checking.

Answer 2

You can live without gdb here - but you have to start solving the problem in a structured manner:

• Don't take anything as granted. Eg, even if you call your program as program > file , don't assume that argv looks the way you assume, but check it by outputting each of them:

   printf("argc: %d\\n", argc); printf("argv[0]: %s\\n", argv[0]); printf("argv[1]: %s\\n", argv[1]); printf("argv[2]: %s\\n", argv[2]); printf("argv[3]: %s\\n", argv[3]); // the se can be expressed better with a for loop - but I'll leave that as an exercise for you 

• Only take things as granted what you have verified: if you know that argc >= 2 , don't access argv[2] and/or argv[3] .

• Don't say

   if(argc<2) { printf("Error"); return 0; } 

  but

   if(argc<2) // according to the point before, better y3 or <4 { printf("Too few command line arguments"); return 1; // not 0; 0 would mean success } 

Answer 3

Joachim Pileborg's answer is obviously correct, just try to run your program as

./a.out a \> b

to prevent the shell from interpreting ">" as a redirection.