Bill Cunningham wrote:
> I have looked this program up and down and I don't see what's wrong
with
> it. But it always breaks and gives me an error "mode error" no matter
which
> mode binary or text I choose. This simple program is supposed to take as
> argv[1] a "b" or "t" for binary or text. It's not taking anything.
>
> #include <stdio.h>
>
> int main(int argc, char **argv) {
> char *b;
> int a;
> FILE *ifp,*ofp;
> if (argc!=4) {
> fprintf(stderr,"usage error\n");
> return -1;
> }
> if (argv[1]=="b") {
> b="rb";
> }
> if (argv[1]=="t") {
> b="rt";
> }
> if (argv[1]!="t"||argv[1]!="b") {
> fprintf(stderr,"mode error\n");
> return -1;
> }
> if ((ifp=fopen(argv[2],b))==0) {
> fprintf(stderr,"open error i\n");
> return -1;
> }
> if ((ofp=fopen(argv[3],b))==0) {
> fprintf(stderr,"open error o\n");
> return -1;
> }
> while(a!=EOF)
> a=fgetc(ifp);
> fputc(a,ofp);
> printf("done\n");
> return 0;}
>
> Is anyone good enough to glance at this and see what's wrong?
>
> Bill
>
>
You can't use == to compare strings. That's what strcmp() is for. This
is the first error I find, not the only one.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
|
