C, How to use exit code properly? - c

Question

When I try to submit my work, the system told me to use the exit code. When I use return 0 and recheck, the system told me to use return 1...
(https://i.stack.imgur.com/dnVLV.png)
:) caesar.c exists.
:) caesar.c compiles.
:( encrypts "a" as "b" using 1 as key
expected "ciphertext: b\...", not "ciphertext: b"
:( encrypts "barfoo" as "yxocll" using 23 as key
expected "ciphertext: yx...", not "ciphertext: yx..."
:( encrypts "BARFOO" as "EDUIRR" using 3 as key
expected "ciphertext: ED...", not "ciphertext: ED..."
:( encrypts "BaRFoo" as "FeVJss" using 4 as key
expected "ciphertext: Fe...", not "ciphertext: Fe..."
:( encrypts "barfoo" as "onesbb" using 65 as key
expected "ciphertext: on...", not "ciphertext: on..."
:( encrypts "world, say hello!" as "iadxp, emk tqxxa!" using 12 as key
expected "ciphertext: ia...", not "ciphertext: is..."
:( handle lack of argv[1]
expected exit code 1, not 0
:( handles non-numeric key
timed out while waiting for program to exit
:( handles too many arguments
expected exit code 1, not 0
How can I fix it and what's wrong with my code?
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
int main(int argc, string argv[])
{
int ok;
char r1;
if (argc == 2)
{
for (int i = 0, s = strlen(argv[1]); i < s; i++)
{
if (!isdigit(argv[1][i]))
{
printf("Sorry\n");
return 0;
}
else
{
ok = atoi(argv[1]);
string c = get_string("Enter:");
printf("ciphertext: ");
for (int t = 0, a = strlen(c); t < a; t++)
{
if (c[t] < 91 && c[t] > 64)
{
r1 = (c[t] - 64 + ok) % 26 + 64;
printf("%c", r1);
}
else if (c[t] < 123 && c[t] > 96)
{
r1 = (c[t] - 96 + ok) % 26 + 96;
printf("%c", r1);
}
else
{
printf("%c", c[t]);
}
}
return 0;
}
}
}
else
{
printf("Sorry\n");
}
printf("\n");
return 0;
}
I try to do well with my homework and all green...

Answer 1

There are multiple issues in your code:
you should return a non zero exit status upon error.
if the number given as a command line argument has more than 1 digit, you perform multiple iterations (one for each digit). You should move the encoding loop out of the first for loop.
using hard coded ASCII values for upper and lower case letters makes the code less portable and hard to read. You should use character constants 'A', 'Z', etc.
r1 = (c[t] - 64 + ok) % 26 + 64; is incorrect and may produce # instead of Z for some inputs. You should use r1 = (c[t] - 65 + ok) % 26 + 65; or better r1 = (c[t] - 'A' + ok) % 26 + 'A';
same mistake for r1 = (c[t] - 96 + ok) % 26 + 96;
Here is a modified version:
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
if (argc != 2) {
fprintf(stderr, "missing argument\n");
return 1;
}
char *arg = argv[1];
char *p;
int shift = (int)strtol(arg, &p, 10);
if (!(arg[0] >= '0' && arg[0] <= '9') || p == arg || *p != '\0') {
fprintf(stderr, "invalid shift argument: %s\n", arg);
return 1;
}
char *s = get_string("Enter string: ");
printf("ciphertext: ");
for (int t = 0; s[t] != '\0'; t++) {
unsigned char c = s[t];
/* assuming ASCII: upper and lowercase letters are contiguous */
if (c >= 'A' && c <= 'Z') {
c = (c - 'A' + shift) % 26 + 'A';
} else
if (c >= 'a' && c <= 'z') {
c = (c - 'a' + shift) % 26 + 'a';
}
putchar(c);
}
putchar('\n');
return 0;
}

Answer 2

You use the exit code by adding a return <value>, at the appropriate code line.
With <value> being what matches your interface definition, in case of your online judge it seems to be a 1.
In your code you at least fail to do so here:
else
{
printf("Sorry\n");
}
which should be
else
{
printf("Sorry\n");
return 1;
}
An alternative is the more explicit https://en.cppreference.com/w/c/program/exit for situations in which the path to the end of the program is not as obvious.
(This is mostly what the comment by Lundin mentions. I turned it into an explicit answer.)
However, to completely satisfy the judge you need to work on your output.
With the info given in the question, a solution for those problems is not possible.