CS50 Caesar program is working but check50 says it isn't - c

I created this program but I'm getting errors on CS50 showing that I didn't do any of it correctly.
The requirements are as follows:
Implement your program in a file called caesar.c in a directory called caesar.
Your program must accept a single command-line argument, a non-negative integer. Let’s call it k for the sake of discussion.
If your program is executed without any command-line arguments or with more than one command-line argument, your program should print an error message of your choice (with printf) and return from main a value of 1 (which tends to signify an error) immediately.
If any of the characters of the command-line argument is not a decimal digit, your program should print the message Usage: ./caesar key and return from main a value of 1.
Do not assume that k will be less than or equal to 26. Your program should work for all non-negative integral values of k less than 2^31 - 26. In other words, you don’t need to worry if your program eventually breaks if the user chooses a value for k that’s too big or almost too big to fit in an int. (Recall that an int can overflow.) But, even if k is greater than 26, alphabetical characters in your program’s input should remain alphabetical characters in your program’s output. For instance, if k is 27,
A should not become [ even though [ is 27 positions away from A in ASCII, per http://www.asciichart.com/[asciichart.com]; A should become B, since B is 27 positions away from A, provided you wrap around from Z to A.
Your program must output plaintext: (without a newline) and then prompt the user for a string of plaintext (using get_string).
Your program must output ciphertext: (without a newline) followed by the plaintext’s corresponding ciphertext, with each alphabetical character in the plaintext “rotated” by k positions; non-alphabetical characters should be outputted unchanged.
Your program must preserve case: capitalized letters, though rotated, must remain capitalized letters; lowercase letters, though rotated, must remain lowercase letters.
After outputting ciphertext, you should print a newline. Your program should then exit by returning 0 from main.
My code:
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(int argc, string argv[])
{
//check if k inputed
if (argc != 2)
{
printf("Usage: ./caesar key\n");
return 1;
}
//value k is the number after ./caesar
int k = atoi(argv[1]) % 26;
int x = 0;
int s = strlen(argv[1]);
//check if k is a positive integer
if (k < 0)
{
printf("Usage: .caesar key\n");
return 1;
}
else
{
//check for arguments
for (int i = 0; i < s; i++)
{
if (isalpha (argv[1][i]))
{
continue;
}
else if (isalnum (argv[1][i]))
{
x++;
}
else
{
continue;
}
}
if (x != s)
{
printf("Usage: ./caesar key\n");
}
else if (x == s)
{
//get plaintext
string plain_text = get_string("plaintext: ");
printf("ciphertext: ");
for (int y = 0; y <= strlen(plain_text); y++)
{
//change letters
if (isalpha(plain_text[y]))
{
char p = plain_text[y];
int cipher_int = p + k;
if (isupper(p))
{
while(cipher_int >= 90)
{
cipher_int -= 26;
}
char cipher_text = cipher_int;
printf("%c", cipher_text);
}
if (islower(p))
{
while(cipher_int >= 122)
{
cipher_int -= 26;
}
char cipher_text = cipher_int;
printf("%c", cipher_text);
}
}
else
{
printf("%c", plain_text[y]);
}
}
printf("\n");
}
}
return 0;
}

It appears that your wrapping is not working correctly. I found that when I used 3 as the key and put "The quick fox jumps over the lazy brown dog." as the plain text, "brown" became "eur`q" when it should be "eurzq". I think you're using >= in your wrapping comparisons when you should use >.

Your check for digits is very cumbersome and does not cause the program to return 1 as required if the argument is incorrect.
Here is a simpler test:
//check for arguments
for (int i = 0; i < s; i++) {
if (!isdigit((unsigned char)argv[1][i])) {
printf("Usage: ./caesar key\n");
return 1;
}
}
Also note that you should stop the encoding loop when the index == the length of the string. therefore the operator should be <.
Another problem is the use of isalpha() and similar functions from <ctype.h> with char values. These functions are undefined for negative values (except EOF). Some platforms define char as signed by default, making isalpha(plaintext[y]) have undefined behavior if the user typed non ASCII text. Cast the argument as (unsigned char) to avoid this problem.
Furthermore, you should not use hardcoded ASCII values such as 90 and 122, use character constants such as 'a' and 'z' for better readability. Doing so would make another error in your encoding loop more obvious: while(cipher_int >= 90) should be if (cipher_int > 'A') and while(cipher_int >= 122) should be if(cipher_int > 'z').
Here is a modified version:
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(int argc, string argv[])
{
// check for a single command line argument
if (argc != 2) {
printf("Usage: ./caesar key\n");
return 1;
}
char *arg = argv[1];
if (*arg == '\0') {
printf("caesar: key cannot be an empty string\n");
return 1;
}
// check that the argument is a non negative number
for (size_t i = 0; arg[i]; i++) {
if (!isdigit((unsigned char)arg[i])) {
printf("Usage: ./caesar key\n");
return 1;
}
}
// value k is the shift number after ./caesar
int k = atoi(argv[1]) % 26;
// get plaintext
string plain_text = get_string("plaintext: ");
printf("ciphertext: ");
for (size_t i = 0; plain_text[i] != '\0'; i++) {
unsigned char c = plain_text[i];
// change letters
if (islower(c)) {
putchar('a' + ((c - 'a') + k) % 26);
} else
if (isupper(c)) {
putchar('A' + ((c - 'A') + k) % 26);
} else {
putchar(c);
}
}
printf("\n");
return 0;
}

Related

Why I am getting an empty value in this implementation of Caesar's cipher?

I've implemented the Caesar's cipher in C, and, despite the algorithm is working, I didn't understood why (sometimes) I get an empty value if I do not subtract the first letter of the alphabet before adding the key. Here's the full code (see line 59 or search for return (letter + k) % 26):
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
string caesar_cipher(string text, int k);
char replace_letter(char letter, int k);
bool is_numeric(string input);
int main(int argc, string argv[])
{
if (argc != 2 || (argc == 2 && !is_numeric(argv[1])))
{
fprintf(stderr, "You must specify a key to the cipher! Exiting...\n");
exit(EXIT_FAILURE);
}
// Convert command line argument to integer.
int k = atoi(argv[1]);
// Prompts user for the text to encrypt
string text = get_string("plaintext: ");
// Returns encrypted text
printf("ciphertext: %s\n", caesar_cipher(text, k));
exit(EXIT_SUCCESS);
}
string caesar_cipher(string text, int k)
{
int text_length = strlen(text);
string ciphered_text = text;
for (int i = 0; text[i] != '\0'; i++)
{
ciphered_text[i] = replace_letter(text[i], k);
}
return ciphered_text;
}
char replace_letter(char letter, int k)
{
// Early return when 'letter' is a non-alphabetical character
if (!isalpha(letter))
{
return letter;
}
char operation_letter = 'a';
if (isupper(letter))
{
operation_letter = 'A';
}
// return (letter + k) % 26; // Sometimes, returns an empty value
return ((letter - operation_letter + k) % 26) + operation_letter;
}
// Loop over characters to check if each one of them is numeric
bool is_numeric(string input)
{
for (int i = 0; input[i] != '\0'; i++)
{
// If character is not numeric
// returns false.
if (isdigit(input[i]) == 0)
{
return false;
}
}
return true;
}
Can anybody explain why this happens?
You need to account for the first letter of the alphabet (either a or A) in your functions because chars are internally represented as an integer number (usually only a single byte, but it depends on the encoding). In ASCII for example, doing a % 26 will result in any of the 26 first values of the ASCII table, none of which are actual letters. Hopefully I made myself clear.

CS50x Caesar - Segmentation fault when inserting isalpha(argv[1])

For the CS50x Problem Caesar I have created a program that encrypts messages using Caesar’s cipher.
For this, the program must recognize that a command-line argument is only a number. So no two or more numbers, no number below zero, and no text.
But as soon as I add the check if it is a text with || isalpha(argv[1]), the program does not work anymore.
The terminal prints the following when I try to run the program:
Segmentation fault
Can anyone tell me what is the problem with the code
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
int main(int argc, string argv[])
{
int kkey = 0;
// Check if correct command-line arguments
if (argc != 2 || atoi(argv[1]) < 0 || isalpha(argv[1])) //segfault here
{
printf("./caesar key\n");
return 1;
}
else
{
kkey = atoi(argv[1]);
}
// Ask for Plaintext to encrypt
string plaintext = get_string("plaintext: ");
for (int i = 0, n = strlen(plaintext); i < n; i++)
{
if (isalpha(plaintext[i]) && islower(plaintext[i]))
{
plaintext[i] = (plaintext[i] - 'a' + kkey) % 26 + 97;
}
else if (isalpha(plaintext[i]) && isupper(plaintext[i]))
{
plaintext[i] = (plaintext[i] - 'A' + kkey) % 26 + 65;
}
printf("%c", plaintext[i]);
}
printf("\n");
return 0;
}
Thank you very much for your help.
As said by #Gerhardh, you can't use strings as argument of isalpha, you need a loop to check each character of the string.
In any case that is not the best approach, using a negated isdigit would be a better option, because it accounts for all the other non numeric characters.
//...
// Check if correct command-line arguments
if (argc != 2 || atoi(argv[1]) < 0)
{
printf("./caesar key\n");
return 1;
}
for(size_t i = 0; i < strlen(argv[1]); i++){
if(!isdigit(argv[1][i])){ //if one of the characters is not a digit 0-9
puts("./caesar key\n");
return 1;
}
}
kkey = atoi(argv[1]); //no else needed
//...
Note that atoi will invoke undefined behavior if the converted value is not representable by an int.
You can use strtol for a more robust alternative.
The link is for Linux man page which I find quite nice, but this is cross-platform.
Again, as stated by #Gerhardh, using character codes may backfire, in this case you are using ASCII encoding, but there are others, this makes your code less portable, use the character instead, 26 + 'a' and 26 + 'A'.

Why am I getting errors?

I'm trying to create a program that accepts cmd line arguments to encipher a plaintext!
The program must accept one cmd line argument after its name when making it and this would be the key which by the plaintext (only) alphabetical characters are rotated by this key (e.g. it's number is added to the real alphabet ASCII number resulting in another alphabet to be printed!
it is supposed to print an error message when one argument is present (e.g. here:/make encipher)
instead of here:/make encipher 12 <-- 12 = key!
I am getting a segmentation fault when running the program without the key argument, why?
This is the full code. I'm posting it because I need to learn where is my fault's exact location
and why is it triggered?!
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h> // To use atoi (converting a string to an int)
#include <ctype.h>
#include <string.h>
bool key_is_numb(string argv[]);
void encipher(string txt, int key);
int main(int argc, string argv[])
{
if (key_is_numb(argv) == false)
{
printf("Usage: ./caesar key\n");
return 1;
}
else
{
int key = atoi(argv[1]);
string plaintext;
if (argc == 2 && key > 0)
{
plaintext = get_string("plaintext: ");
encipher(plaintext, key); // A function that prints the ciphered text
return 0; // returns Zero as main return value which means "All good"!
}
else if (argc == 1 || argc > 2 || key <= 0)
{
printf("Usage: ./caesar key\n");
return 1;
}
} // End else.
} // End main()å func.
bool key_is_numb(string argv[])
{
int n = strlen(argv[1]);
for (int i = 0; i < n; i++) // checking element by element in the second string of the argv[] array of strings
{
if (isdigit(argv[1][i]) == 0) // if the entered string "key" contains chars other than digits.
{
return false; // break out of the if statement & the entire function key_is_numb()
// and return false as soon as a letter is encountered.
}
else
{
continue; // go up & start the next iteration for the for loop.
}
// if only digits encountered then this for loop will come to an end and exist from here.
} // End for loop
return true; // function exits and return boolean true from here.
} // End key_is_numb() func.
void encipher(string txt, int key)
{
printf("ciphertext: ");
for (int i = 0, n = strlen(txt); i <= n; i++) // strlen counts the number of elements in a string excluding '\0'
{
char c = txt[i];
if (isalpha(c))
{
if (isupper(c))
{
char m = 'A'; // This is a modifyer character equals to 'A' = 65 so that it is indexed # ZERO!
printf("%c", (c - m + key) % 26 + m );
//c = ((((int)txt[i] - 65) + key) % 26) + 65; // char c = 65 <-- 65 is an ASCII code equals 'A'
}
else if (islower(c))
{
char m = 'a'; // This is a modifying character 'a' = 97
printf("%c", (c - m + key) % 26 + m );
}
}// End if(alpha).
else
{
printf("%c", c);
}
} // End for().
printf("\n");
} // End encipher() func.
int n = strlen(argv[1]);
in key_is_numb() and
int key = atoi(argv[1]);
in main().
If you didn't enter a key argument, argv[1] as equal as argv[argc] is a null pointer as stated in C17, §5.1.2.2.1/2.
Any attempt to access its data is undefined behavior and probably caused the segmentation fault.
Well you are assuming that argv[1] is defined in key_is_numb. However, in C and C++, the second parameter of the main function contains command line arguments. Which, in your case will be the name of the binary as the first element, then any other arguments. This is why when you are running the program without arguments, it will segfault, as there are no argument to put in argv, and no default value either.
You should always check the size of argv, by using the number stored in argc, before trying to read anything in argv.
Your segmentation fault comes from this line int n = strlen(argv[1]);, but I'd highly suggest you to learn to use debugger software like valgrind, which if the program has been compiled with debug flag will tell you the exact line.
Other debugger are really useful too, so you should learn to use them, as they usually report this kind of errors.
Your code asumes there is always an argv[1]. You should check argc which tells the number of arguments. For example:
int main(int argc, string argv[])
{
if (argc < 2) {
printf("Key required\n");
exit (1);
}

C program doesnt handle non numeric key

Good evening people hope yall are doing well, i come to you all because i need some help regarding some C code i wrote, keep in mind that i just started coding in C so sorry if this question sounds dumb.
Basically I am doing the CS50 and we are writing a program that encripts messages so we first require the user for a command line argument which will be the key we will use to transform the plain text into cipher text. so basically if the user runs the command with lets say a 2 i.e: ./caesar 2 all the words he will type later will be "run" two sides.
my program works how is supposed and if the user types a letter instead of a number the program wont prompt for the users message. however, if the user runs the command line like this for example, ./caesar 8x the program will run eventhough the user typed a letter so any ideas on how to iterate thru the user command argument and if there is a letter the program should run? thanks!
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <cs50.h>
int main(int argc, string argv[])
{
if (argc == 2 && isdigit(*argv[1]))
{
int key = atoi(argv[1]);
string plaintext = get_string("plaintext: ");
printf("ciphertext: ");
for (int i = 0, n = strlen(plaintext); i < n; i++)
{
if (isupper(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 65) % 26) + 65);
}
else if (islower(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 97) % 26) + 97);
}
else
{
printf("%c", plaintext[i]);
}
}
printf("\n");
return 0;
}
else if (argc == 1)
{
printf("NO KEY\n");
return 1;
}
else if (argc >= 3 || argv[1] == (string) argv[1])
{
printf ("Usage: ./caesar key\n");
return 1;
}
}
EDIT. SOLVED
this is how the code ended up looking, thanks #bruno for the help btw
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <cs50.h>
int main(int argc, string argv[])
{
if (argc == 2)
{
char *endp;
long lkey;
int errno = 0;
lkey = strtol(argv[1], &endp, 10);
if ((errno != 0) || (*argv[1] == 0) || (*endp != 0) || (lkey < 0) || (((int) lkey) != lkey) || argc >= 3)
// this statement check to see the characters on argv[1] are all digits
{
printf("Usage: ./caesar key\n");
}
else // if the argv[1] is all digits then prompts the user for a plaintext
{
int key = atoi(argv[1]); // converts the key into an integer
string plaintext = get_string("plaintext: ");
printf("ciphertext: ");
for (int i = 0, n = strlen(plaintext); i < n; i++)
// goes thru each of the chars in plaintext and determines if is uppercase, lowercase or none.
{
if (isupper(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 65) % 26) + 65);
// it takes 65 and then sums it back to convert the character from the ASCII uppercase index and back
}
else if (islower(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 97) % 26) + 97);
// if the case is lower it takes 97 and then adds 97 back just to maintain the ASCII index.
}
else // if it not a lower case nor an uppercase, which means is a symbol then leave it like that.
{
printf("%c", plaintext[i]);
}
}
printf("\n");
return 0;
}
}
else if (argc == 1) // if the user doesnt prompt a key print no key to user
{
printf("NO KEY\n");
return 1;
}
else if (argc >= 3) // if user prompts 3 or more keys into argv then prints error message regargind the usage
{
printf("Usage: ./caesar key\n");
return 1;
}
}
if the user runs the command line like this for example, ./caesar 8x the program will run eventhough the user typed a letter
you get the number from the program argument using atoi, it stops when a non-digit is reached, so the result is the same for 8 and 8x.
If the argument is not compatible with a number since the beginning, atoi will silently return 0, this is why atoi is dangerous. Anyway here you cannot be in that case thank to your test isdigit(*argv[1]) before.
any ideas on how to iterate thru the user command argument and if there is a letter the program should run
Probably you wanted to say the program should not run.
To check if all the argument is compatible with a number you can use strtol to do the conversion :
if (argc == 2) {
char * endp;
long lkey;
errno = 0;
lkey = strtol(argv[1], &endp, 10);
if ((errno != 0) || (*argv[1] == 0) || (*endp != 0) ||
(lkey < 0) || (((int) lkey) != lkey)) {
printf("invalid argument %s\n", argv[1]);
}
else {
int key = (int) lkey;
...
Because you currently use an int for the key I check if the value is compatible with an int, I also check it is positive because this is compatible with your use and in your version you verify the first character is a digit
Good evening,
I don't know if I understood your question right, but I will give it a try.
The problem is that you are checking if the args counter is 2, and the 2nd argument is a digit. if so, do the following code you have written. but if not?
You are not handling the case of argc==2 and argv[1] is not a digit.

How Do I Rotate A Char Alphabetically By Integer X?

I've been trying different solutions but am not sure where to look for the solution.
I prompt the user for "plaintext" and whatever their input is, the chars in their input need to be rotated alphabetically by a number (aka the key) which they provide.
For example: plaintext: HELLO would spit out ciphertext: IFMMP if the key were 1.
Assuming the key will always be a number, here is what my code looks like which attempts to rotate each char by Key: 1. I'm a real noob so please break it down is possible.
{
string s = get_string("plaintext: ");
printf("ciphertext: %s\n", s + 1);
}
The remaining code (which includes identifying and filtering out the key is:
#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main(int argc, string argv[])
{
int counter = 0;
if (argc == 2)
{
for(int k = 0; k <strlen(argv[1]); k++)
{
if (isdigit(argv[1][k]))
{
counter++;
}
}
if (strlen(argv[1]) == counter)
{
string s = get_string("plaintext: ");
if(s)
{
printf("ciphertext: %s\n", s + 1);
free(s);
}
}
else
{
printf("Usage: ./caesar key\n");
}
}
else
{
printf("Usage: ./caesar key\n");
}
}
Any assistance would be appreciated.
I do not have cs50, so cannot test everything in your code, but addressing your primary question: How Do I Rotate A Char Alphabetically By Integer X, the code section you have identified as where the rotation occurs:
if (strlen(argv[1]) == counter)
{
string s = get_string("plaintext: ");
printf("ciphertext: %s\n", s + 1);
}
But this section doesn't really rotate the text. It rather attempts to print the string obtained from the call to get_string after adding 1 to it. First, this is not modifying s at all. Second, s + 1 is not a legal operation. To rotate s, additional code is needed between those two lines.
If I understand get_string(), it is essentially a combination of printf(), fgets(,,stdin) and strdup().
So after the call to that function you are left with s which will be populated with whatever the user typed into stdin. For illustration, let
say s contains "what the user typed". How that string is represented in memory, including the NULL terminator, can be depicted as follows:
|w|h|a|t| |t|h|e| |u|s|e|r| |t|y|p|e|d|0|
To rotate this by X, as you have stated, each character needs to be modified by adding the value X, in this case 1
Showing ASCII equivalent values:
|119|104|97|116|32|116|104|101|32|117|115|101|114|32|116|121|112|101|100|0|
The rotated string then would be each value + 1:
For the general case, assume rotation value is read in from command line argument and placed into int n = 0;
earlier in code:
if(isdigit(argv[2][0]))
{
n = atoi(argv[2])
}
The rotation: (the following is complete. It may need a little debugging around the corner case [where values wrap around the end of ASCII values when added to n.] Lines are commented indicating where I used ANSI C instead of CS50. [I do not have CS50])
int main(int argc, char *argv[])//note: command line requires a numeric
// argument be used: prog.exe 12 (for n == 12)
{
//char s[80];//used to test (I do not have cs50)
char *rotated = NULL;
int i;
if(argc != 2)
{
printf("Usage: prog.exe <positive int value.>\nExiting\n");
return 0;
}
if(isdigit(argv[1][0]) == 0)
{
printf("Usage: prog.exe <positive int value.>\nExiting\n");
return 0;
}
int n = atoi(argv[1]);
string s = get_string("plaintext: ");
// printf("Enter text to be rotated\n");//used to test (I do not have cs50)
// scanf("%[^\n]", s);
// string rotated = strdup(s); //preserve s by manipulating identical string
rotated = StrDup(s);
if(rotated)
{ //advance value by n
int origVal = 0;
int len = strlen(rotated);
for(i=0;i<len;i++)
{
origVal = rotated[i];
if((rotated[i] + n) <= 127)
{
rotated[i] += n;
}
else
{
rotated[i] = abs(127 - (origVal + n));
if(rotated[i] == 0) rotated[i] += 23; //skip to printable characters (http://facweb.cs.depaul.edu/sjost/it212/documents/ascii-pr.htm)
//if(rotated[i] == 0) rotated[i] += 1; //skip only ASCII NULL value
}
}
rotated[i] = 0; //adding NULL to end of string
printf("ciphertext: %s\n", rotated);
free(rotated);
}
return 0;
}
You can achieve this with typecasting:
char input = 'f';
char output = (char)((int)input + 1);
You should also handle edge cases, i.e. when the character value is near the end of the alphabet.

Resources