fopen 的 Linux 分段错误

Question

Does anyone know what's wrong with this code? I keep getting segmentation fault

int main (int argc, char **argv)
{
    FILE *in, *out;
    in = fopen(argv[1],"r");
    out = fopen(argv[2],"w");
    fseek(in,0,SEEK_END);
    ...
    fseek(in,0,SEEK_SET);

I did ./a.out filename1 filename2

I tried copying the arguments into string variables and I didn't have any problems

char f1[100],f2[100];
strcpy(f1,argv[1]);
strcpy(f2,argv[2]);
FILE *in, *out;
in = fopen(f1,"r");
out = fopen(f2,"w");

Answer 1

Does anyone know what's wrong with this code?

You don't have any error checking code. You assume the calls to fopen are successful.

in = fopen(argv[1],"r");
if ( in == NULL )
{
   // Problem opening the file.
   // Print the cause of the problem and exit.
   perror("Unable to open the file");
   exit(EXIT_FAILURE);
}

Add similar code for out .