I'm trying to compile this program and if I keep the statements:
if (argc !=2 )
syntax(argv[0]);
I get parse errors. I can not figure out where the parse error is. I'm
sure I'm overlooking something. Can anyone here
see it?
-- Cut here ---
#include
#include
#include
#include
void syntax(char *progname); /* Syntax Display Function */
void main(int argc, char *argv[])
{
if(argc != 2) /* I get syntax errors If I use this if */
syntax(argv[0]); /* I get syntax error if I use this */
struct stat cstat;
ino_t myino;
if(stat(argv[1],&cstat)== -1)
{
fprintf(stderr,"error opening file.\n");
syntax(argv[0]);
}
myino=cstat.st_ino;
fprintf(stdout,"%s: %ld\n",argv[1],myino);
}
void syntax(char *progname)
{
fprintf(stderr,"Usage: %s:\n",progname);
fprintf(stderr,"--------------------------\n");
fprintf(stderr,"%s [FILENAME]\n",progname);
fprintf(stderr,"\nFILENAME=filename to get inode location\n");
exit(1);
}
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