error: expected '(' before 'else'

Hello :)

I am trying to write a program that decrypts a string that the user enters with the caesar cypher.

I am pretty sure the rest of my program is correct but I am getting this error: error: Expected '(' before 'else' for line 37 (if else (plain [i] > 96 && plain[i] < 123)

here is the program:

Code:

---

/* The program encrypts a string with the caesar cypher*/
#include<cs50.h>
#include<stdio.h>
#include<ctype.h>
#include<stdlib.h>
#include<string.h>

int
main(int argc, char *argv[])
{
      // check if the argument entered is a digit
      if (!isdigit(*argv[1]) || argc !=2)
      {
            printf("Please enter one number as the argument to use to encrypt\m")
            return 1;
      }
      // convert the argument to an int
      int k = atoi(argv[1]);
      // get the user's plaintext to encrypt
      printf("\nplaintext:  ");
      char *plain = GetString();
      // encrypt
      printf("\cyphertext:  ");
      for (int i = 0, n = strlen(plain); i < n; i++)
      {
            // if the letter is lowercase
            if(plain[i] > 64 && plain[i] < 91)
            {
                  int ascii = plain[i];
                  int p = (ascii - 65);
                  int c = (p + k) % 26;
                  int cypher_p = (c + 65);
                  printf("%c", cypher_p);
            }
            // if the letter is uppercase
            // this is where the error is
            if else(plain[i] > 96 && plain[i] < 123)
            {
                  int ascii = plain[i];
                  int p = (ascii - 97);
                  int c = (p + k) % 26;
                  int cypher_p = (c + 97);
                  printf("%c", cypher_p);
            }
            // if the element in plain is a space print a space
            if else(plain [i] == " ")
            {
                  printf(" ");
            }
      }
      printf("\n");
}

---

Why do I keep getting that error?

I literally just started programming like two weeks ago so please explain the fix (unless it is just something really stupid that I'm missing haha).

You need to change "if else" to "else if".

Code:

---

          // if the letter is lowercase
            if(plain[i] > 64 && plain[i] < 91)
            {
                  int ascii = plain[i];
                  int p = (ascii - 65);
                  int c = (p + k) % 26;
                  int cypher_p = (c + 65);
                  printf("%c", cypher_p);
            }
          if else(plain[i] > 96 && plain[i] < 123)  <=== lose the if
          {
                  int ascii = plain[i];
                  int p = (ascii - 97);
                  int c = (p + k) % 26;
                  int cypher_p = (c + 97);
                  printf("%c", cypher_p);

---

Wow I knew it was something stupid

Thanks haha, I have been tearing my hair out over this.

Quote:

---

Originally Posted by BigWigly

Wow I knew it was something stupid

Thanks haha, I have been tearing my hair out over this.

---

Actually the correct grammar there would be ... else if ...

Actually now I am getting an error for the line:
else if (plain [i] == " ")
printf(" ");

It should be pretty obvious what I want to do.... How do I do it?? haha

You can't compare strings with ==, you need to use strcmp. If you want to check a single char, use single quotes: ' '.

How can I use strcmp() with plain[i] and ' '?

I'm guessing I would check if it == 0?

I tried

Code:

---

else if(strcmp(plain[i], ' ') == 0)

---

didn't work :'(

Thanks a lot for the really fast responses by the way :)

Sorry, I should have been more clear. It looks like you want to check against a single space character. Single quotes ' ' are for single characters, double quotes " " are for strings. You can compare single characters with ==, so you want if (plain[i] == ' '). If you wanted to compare strings, you would need if (strcmp(some_string, "foobar") == 0). Hope that makes a little more sense.

YESSSSSSS

Thank you so much everyone :D I finally got it to work!

You're welcome!