Copying one file to another: scanf keeps looping - c

Question

I am creating a program that takes two command-line arguments: the first line argument is the name of the file to be copied and the second is the new file. If the second argument is missing, copy the file to stdout. If both arguments are missing, the program should read from stdin and print to stdout ie. ./a.out input.txt output.txt
I did the following but I'm facing a problem where scanf keeps looping and does not quit:
int main(int argc, char *argv[]) `
{
char text[10];
FILE *input;
FILE *output;
char ch;
printf("%s", argv[0]);
if (argc == 3)
{
input = fopen(argv[1], "r");
output = fopen(argv[2], "w");
while ((ch = fgetc(input)) != EOF)
{
fputc(ch, output);
ch = fgetc(output);
}
}
if (argc == 2)
{
input = fopen(argv[1], "r");
while ((ch = fgetc(input)) != EOF)
{
printf("%c", ch);
}
printf("\n");
}
if (argc == 1)
{
scanf("%c", text);
// here it keeps looping
}
fclose(input);
return 0;
}
Screen shot with cursor at end of line after some input before return is hit

Answer 1

It's not looping, it's waiting for input. You've requested scanf("%c"), a character, but scanf won't continue until you press enter. I think you meant scanf("%s") or getch(), but it isn't clear.
It will throw an error on the next line as fclose() is closing a FILE* that isn't initialized in the case of only 1 argument.
You could probably rework this homework example to use fopen() on STDIN/STDOUT in the case of missing parameters so that you aren't writing redundant code.

Answer 2

There are multiple problems in your code:
You output argv[0] to stdout: if argc < 3, you are supposed to copy to stdout, this extra output is corrupting the output. You might instead output to stderr or remove this line completely.
you do not check for fopen failure, causing undefined behavior if either file cannot be opened.
ch has type char which is too small to accommodate for all return values of fgetc(). On machines with 8-bit bytes, fgetc() has 257 possible return values, you must make ch an int to reliably distinguish EOF from all other return values.
in the fgetc() / fputc() loop, you read 2 bytes in each iteration of the loop but only write one byte.
the scanf() on the last case is simply waiting for input as you are supposed to copy from stdin to stdout, but the program will stop as soon as you hit the Enter key. You should just use the same fgetc()/fputc() loop for all copying loops.
you should include <stdio.h>
Here is a modified version:
int main(int argc, char *argv[]) {
FILE *input = stdin;
FILE *output = stdout;
int ch;
if (argc > 1) {
input = fopen(argv[1], "r");
if (input == NULL) {
fprintf("cannot open input file %s\n", argv[1]);
return 1;
}
}
if (argc > 2) {
output = fopen(argv[2], "w");
if (output == NULL) {
fprintf("cannot open output file %s\n", argv[2]);
return 1;
}
}
while ((ch = fgetc(input)) != EOF) {
fputc(ch, output);
}
if (argc > 2) {
fclose(output);
}
if (argc > 1) {
fclose(input);
}
return 0;
}