Question

How can I convert a character into an integer value? For example, I’ve been trying to read “A” as 1, “B” as 2, and so forth. I tried comparing the character to each letter in the alphabet and return the appropriate value.

int intvalue(char letter)
{
if(letter == "A")
    return 1;
else if(letter == "B")
    return 2;
else if(letter == "C")
    return 3;
else if(letter == "D")
    return 4;
else if(letter == "E")
    return 5;
else if(letter == "F")
    return 6;
else if(letter == "G")
    return 7;
else if(letter == "H")
    return 8;
else if(letter == "I")
    return 9;
else if(letter == "J")
    return 10;
else if(letter == "K")
    return 11;
else if(letter == "L")
    return 12;
else if(letter == "M")
    return 13;
else if(letter == "N")
    return 14;
else if(letter == "O")
    return 15;
else if(letter == "P")
    return 16;
else if(letter == "Q")
    return 17;
else if(letter == "R")
    return 18;
else if(letter == "S")
    return 19;
else if(letter == "T")
    return 20;
else if(letter == "U")
    return 21;
else if(letter == "V")
    return 22;
else if(letter == "W")
    return 23;
else if(letter == "X")
    return 24;
else if(letter == "Y")
    return 25;
else if(letter == "Z")
    return 26;

}

I got “error: ISO C++ forbids comparison between pointer and integer”. Does anyone know how to fix this? Or even better, a different way to go about this? I feel like my above function is very brute-forceish.

Answer 1

You need to use character literals, not string literals, e.g.,

if (letter == 'A')
              ^ note the single quotes

That said, if you are willing to assume you are running on a system using the ASCII character set, you can simply use arithmetic:

int charIndex = (letter - 'A') + 1;

In the ASCII character set, letters are at consecutive indices, so this works. This may not work with other character sets.