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Finding Capital Letter

GoranGaming (128)
I've made a function that finds the capital or lowercase letter of a letter that was entered, but I find my function too long, and I wonder if there is any shorter and easier way to do this than I did.

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char hittaStorLitenBokstav(char test){
    char s;
    switch(test){
        case 'a':
            s = 'A';
            break;
        case 'b':
            s = 'B';
            break;
        case 'c':
            s = 'C';
            break;
        case 'd':
            s = 'D';
            break;
        case 'e':
            s = 'E';
            break;
        case 'f':
            s = 'F';
            break;
        case 'g':
            s = 'G';
            break;
        case 'h':
            s = 'H';
            break;
        case 'i':
            s = 'I';
            break;
        case 'j':
            s = 'J';
            break;
        case 'k':
            s = 'K';
            break;
        case 'l':
            s = 'L';
            break;
        case 'm':
            s = 'M';
            break;
        case 'n':
            s = 'N';
            break;
        case 'o':
            s = 'O';
            break;
        case 'p':
            s = 'P';
            break;
        case 'q':
            s = 'Q';
            break;
        case 'r':
            s = 'R';
            break;
        case 's':
            s = 'S';
            break;
        case 't':
            s = 'T';
            break;
        case 'u':
            s = 'U';
            break;
        case 'v':
            s = 'V';
            break;
        case 'x':
            s = 'X';
            break;
        case 'y':
            s = 'Y';
            break;
        case 'z':
            s = 'Z';
            break;
        case 'A':
            s = 'a';
            break;
        case 'B':
            s = 'b';
            break;
        case 'C':
            s = 'c';
            break;
        case 'D':
            s = 'd';
            break;
        case 'E':
            s = 'e';
            break;
        case 'F':
            s = 'f';
            break;
        case 'G':
            s = 'g';
            break;
        case 'H':
            s = 'h';
            break;
        case 'I':
            s = 'i';
            break;
        case 'J':
            s = 'j';
            break;
        case 'K':
            s = 'k';
            break;
        case 'L':
            s = 'l';
            break;
        case 'M':
            s = 'm';
            break;
        case 'N':
            s = 'n';
            break;
        case 'O':
            s = 'o';
            break;
        case 'P':
            s = 'p';
            break;
        case 'Q':
            s = 'q';
            break;
        case 'R':
            s = 'r';
            break;
        case 'S':
            s = 's';
            break;
        case 'T':
            s = 't';
            break;
        case 'U':
            s = 'u';
            break;
        case 'V':
            s = 'v';
            break;
        case 'X':
            s = 'x';
            break;
        case 'Y':
            s = 'y';
            break;
        case 'Z':
            s = 'z';
    }
    return s;
}
GoranGaming (128)
Don't worry, I came up with a remarkably shorter one myself :)

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char hittaStorLitenBokstav(char test){
    char s;
    char alphabet1[26] = "abcdefghjklmnopqrstuvwxyz";
    char alphabet2[26] = "ABCDEFGHJKLMNOPQRSTUVWXYZ";
    for(int x=0;x<26;x++){
        if(test == alphabet1[x]){
            s = alphabet2[x];
        }
        else if(test == alphabet2[x]){
            s = alphabet1[x];
        }
    }
    return s;
}
Last edited on
Disch (13742)
char alphabet1[26] = "abcdefghjklmnopqrstuvwxyz";

You'd need 27 characters. 26 for the alphabet + 1 for the null terminator.

Also, you are missing i / I.

Also, the output is undefined if 'test' is not an alphabetical character, since you don't initialize 's'


Anyway, here's another way to do it:

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char hittaStorLitenBokstav(char test)
{
    if( (test >= 'A') && (test <= 'Z') )
        return test - 'A' + 'a';
    if( (test >= 'a') && (test <= 'z') )
        return test - 'a' + 'A';

    return test;
}
Momoironeko (13)
Another simple way to do it using for loops.

Code 
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char hittaStorLitenBokstav(char test)
{
     for (int jjj = 0; jjj < 26; ++jjj)
          if (test == 'a' + jjj)
               test = 'A' + jjj;
     return test;
}


Explanation
Loop will run 26 times. Each time adding 1 to jjj starting at zero and ending at 25.

'a' + 0 == 'a'
'A' + 0 == 'A'

'a' + 1 == 'b'
'A' + 1 == 'B'

'a' + 25 == 'z'
'A' + 25 == 'Z'


Read
http://www.cplusplus.com/doc/ascii/
http://www.cplusplus.com/doc/tutorial/control/
Last edited on
TheIdeasMan (6817)
And another way is to use the toupper function - not sure whether you are allowed to use this, or were unaware of it.

http://www.cplusplus.com/reference/cctype/toupper/


This is in addition to Disch's post which is great for showing the OP how to make use of the char values.

HTH

Edit: there are 2 other versions of this function - which you can find in the reference. The 1 quoted is the C version. Here is a C++ one :

http://www.cplusplus.com/reference/locale/toupper/
Last edited on
Duthomhas (13212)
GoranGaming (128)
@ Disch

I don't really get how your function works, but it certainly does.

@ TheIdeasMan

Actually I didn't know about it :)
Paoletti301 (315)
@goran
@ Disch

I don't really get how your function works, but it certainly does


it works because each character has a value associated with it, so 'a' = 97, 'b' = 98, and so on.

capitals have values with them as well, 'A' = 65, 'B' = 66, and so on.

his code checks whether the input is a capital or not and then uses the values of the characters to do the conversion

so "C" for example

'C' - 'A' + 'a' = 'c' . . . which is the same as 67 - 65 + 97 = 99 or 'c'

i hope this helps explain whats going on there :D

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char test = 'a';
int teste = test;
std::cout << teste;


you can use that to look at character values if you want

EDIT:spelling
Last edited on
GoranGaming (128)
Now I get it :) Thank you for your explanation.
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