cdo-whilelogical-operatorslogical-orlogical-and

I don't know what I'm doing wrong. I think my condition is right


char ch;
    do
    {
        printf("Digite aqui um caractere: ");
        scanf(" %c", &ch);
    } while ((ch < 'A' && ch > 'Z' ) || (ch < 'a' && ch > 'z') || ch != '.');
    return ch;

I tried all sort of things on this condition and I can't make it happen. I want to return the value of "ch" when the input is [A-Z] or [a-z] or '.'.


Solution

  • In the condition of the loop

    while ((ch < 'A' && ch > 'Z' ) || (ch < 'a' && ch > 'z') || ch != '.');
    

    for example thus subexpression

    (ch < 'A' && ch > 'Z' )
    

    is always evaluates to logical false because a character can not be at the same time less than 'A' and greater than 'Z'.

    To simplify the condition at first rewrite it for the case when the loop should be interrupted.

    The loop is interrupted when

    ( ( 'A' <= ch && ch <= 'Z' ) || ( 'a' <= ch && ch <= 'z') || ( ch == '.' ));
    

    Now write its negation

    !( ( 'A' <= ch && ch <= 'Z' ) || ( 'a' <= ch && ch <= 'z') || ( ch == '.' ));
    

    You will get

    ( !( 'A' <= ch && ch <= 'Z' ) && !( 'a' <= ch && ch <= 'z') && !( ch == '.' ));
    

    It is the same as

    ( ( !('A' <= ch ) || !( ch <= 'Z' ) ) && ( !( 'a' <= ch ) || !(ch <= 'z' )) && ( ch != '.' ));
    

    or

    ( ( ch < 'A' || ch > 'Z' ) && ( ch < 'a' || ch > 'z' ) && ( ch != '.' ));
    

    So you will have

    while ( ( ch < 'A' || ch > 'Z' ) && ( ch < 'a' || ch > 'z' ) && ( ch != '.' ));
    

    Another approach is to use standard function tolower or toupper declared in the header <ctype.h> to simplify the condition as for example

    while ( ( toupper( ( unsigned char )ch ) < 'A' || toupper( unsigned char )ch ) > 'Z' ) && ( ch != '.' ));
    

    Or according to the remarkable idea of @Gerhardh you can also write

    while ( !isalpha( ( unsigned char )ch ) && ( ch != '.' ));