How do I increment letters in c++? - c++

Question

I'm creating a Caesar Cipher in c++ and i can't figure out how to increment a letter.
I need to increment the letter by 1 each time and return the next letter in the alphabet. Something like the following to add 1 to 'a' and return 'b'.
char letter[] = "a";
cout << letter[0] +1;

Answer 1

This snippet should get you started. letter is a char and not an array of chars nor a string.
The static_cast ensures the result of 'a' + 1 is treated as a char.
> cat caesar.cpp
#include <iostream>
int main()
{
char letter = 'a';
std::cout << static_cast<char>(letter + 1) << std::endl;
}
> g++ caesar.cpp -o caesar
> ./caesar
b
Watch out when you get to 'z' (or 'Z'!) and good luck!

Answer 2

It works as-is, but because the addition promotes the expression to int you want to cast it back to char again so that your IOStream renders it as a character rather than a number:
int main() {
char letter[] = "a";
cout << static_cast<char>(letter[0] + 1);
}
Output: b
Also add wrap-around logic (so that when letter[0] is z, you set to a rather than incrementing), and consider case.

Answer 3

You can use 'a'+((letter - 'a'+n)%26);
assuming after 'z' you need 'a' i.e. 'z'+1='a'
#include <iostream>
using namespace std;
int main()
{
char letter='z';
cout<<(char)('a' + ((letter - 'a' + 1) % 26));
return 0;
}
See this https://stackoverflow.com/a/6171969/8511215

Answer 4

Does letter++ work?
All in all char is a numeric type, so it will increment the ascii code.
But I believe it must be defined as char letter not an array. But beware of adding one to 'Z'. You will get '[' =P
#include <iostream>
int main () {
char a = 'a';
a++;
std::cout << a;
}
This seems to work well ;)

Answer 5

char letter = 'a';
cout << ++letter;

Answer 6

waleed#waleed-P17SM-A:~$ nano Good_morning_encryption.cpp
waleed#waleed-P17SM-A:~$ g++ Good_morning_encryption.cpp -o Good_morning_encryption.out
waleed#waleed-P17SM-A:~$ ./Good_morning_encryption.out
Enter your text:waleed
Encrypted text:
jnyrrq
waleed#waleed-P17SM-A:~$ cat Good_morning_encryption.cpp
#include <iostream>
#include <string>
using namespace std;
int main() {
//the string that holds the user input
string text;
//x for the first counter than makes it keeps looping until it encrypts the user input
//len holds the value (int) of the length of the user input ( including spaces)
int x, len;
//simple console output
cout << "Enter your text:";
//gets the user input ( including spaces and saves it to the variable text
getline(cin, text);
//give the variable len the value of the user input length
len = (int)text.length();
//counter that makes it keep looping until it "encrypts" all of the user input (that's why it keeps looping while its less than len
for(x = 0; x < len; x++) {
//checks each letts (and spaces) in the user input (x is the number of the offset keep in mind that it starts from 0 and for example text[x] if the user input was waleed would be w since its text[0]
if (isalpha(text[x])) {
//converts each letter to small letter ( even though it can be done another way by making the check like this if (text[x] =='z' || text[x] == 'Z')
text[x] = tolower(text[x]);
//another counter that loops 13 times
for (int counter = 0; counter < 13; counter++) {
//it checks if the letts text[x] is z and if it is it will make it a
if (text[x] == 'z') {
text[x] = 'a';
}
//if its not z it will keeps increamenting (using the loop 13 times)
else {
text[x]++;
}
}
}
}
//prints out the final value of text
cout << "Encrypted text:\n" << text << endl;
//return 0 (because the the main function is an int so it must return an integer value
return 0;
}
Note: this is called caeser cipher encryption it works like this :
ABCDEFGHIJKLMNOPQRSTUVWXYZ
NOPQRSTUVWXYZABCDEFGHIJKLM
so for example my name is waleed
it will be written as : JNYRRQ
so its simply add 13 letters to each letter
i hope that helped you

Answer 7

It works but don't forget that if you increment 'z' you need to get 'a' so maybe you should pass by a check function that output 'a' when you get 'z'.

Answer 8

cast letter[n] to byte* and increase its referenced value by 1.