Thread: Big Letter became small letter

I writted the following code:

Code:

#include <stdio.h>

main()

{

        char ch, newch;

        printf("Add a character: ");
        scanf("%c", &ch);
        newch =  223 & ch;
        printf("Big letter: %c\n", newch);

}

so when i run it i write a small letter for example:

Add a character: a
Big letter: A

now, i try create a program that will get a big letter and make it small but i don't have no idea how to do this in ASCII I tried using
newch= ch & 223; but still the same?
any ideas could be useful, thanks!

You do know you can just use standard functions, guarinteed to work on your current character set, to do this for you? toupper and tolower

Quzah.

what is touper and tolower,
i hear them for first time?

I read the manual of C and writed this:
#include <stdio.h>
main()
{
char c;
printf("Add a character: ");
scanf("%c", &c);
toupper(c);
printf("The character is %c\n");
}

but when running and writing a charachet i get
the character is 7
help me please

clicky click

Quzah.

yes, i read the man from my unix box but i don't get why it has it as an integer as it should be char, saw my program?

THe ascii value of a character is an Integer.
Airgo, any single character can be represented by the appropriate related ascii character value.

www.asciitable.com best web site ever for all your ascii number needs.

I change all the %c to %i and the char to int and now get:
The character is -1073743308

The format specifier for a character, using the *printf and *scanf functions is in fact %c. However, those functions mentinoed take integers because for one thing, you can't represent EOF in a char.

Quzah.

Ok,
i change drthe code to this:
#include <stdio.h>
main()
{
int c;
printf("Add a character: ");
scanf("%c", &c);
toupper(c);
printf("The character is %c\n");
}

but still getting
the carachter is 4
???

Where is the ch salem?
shoudnlt' i say char c, ch; ???