Printable View
I writted the following code:
so when i run it i write a small letter for example:Code:#include <stdio.h>
main()
{
char ch, newch;
printf("Add a character: ");
scanf("%c", &ch);
newch = 223 & ch;
printf("Big letter: %c\n", newch);
}
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
???
Code:char c;
scanf( "%c", &ch );
ch = toupper( ch );
Where is the ch salem?
shoudnlt' i say char c, ch; ???
It was mostly to show you how to use toupper()
I think toupper() and tolower() are in ctype.h so remember to include it.
If that was the problem, he wouldn't be getting 7 and 4.Quote:
THe ascii value of a character is an Integer.
Airgo, any single character can be represented by the appropriate related ascii character value.
If that was the problem, he'd have gotten a compile error.Quote:
I think toupper() and tolower() are in ctype.h so remember to include it.
I don't mean to cut people down, I'm just trying to stop thing from getting more confusing.
You must give printf something to print:Quote:
Originally Posted by cogeek
Regards.Code:printf("The character is %c\n", c);
Dave
OK: i writed this:
#include <stdio.h>
main()
{
int c, ch;
printf("Add a character: ");
scanf("%c", &ch);
ch = tolower(ch);
printf("The character is %c\n", ch);
}
but when riting a character i get the same for example:
bash-2.05b$ ./big
Add a character: A
The character is A
The biggest problem is with scanf. It expects a pointer to char, but you gave it a pointer to int. That's not a good thing.
Code:#include <stdio.h>
#include <ctype.h>
int main(void)
{
char ch;
printf("Add a character: ");
scanf("%c", &ch);
ch = (char)tolower(ch);
printf("The character is %c\n", ch);
return 0;
}
I din't get what the (char) mean
?
Casts the returned integer down to a char. Both are integral types.
Look in your book for casting
>I din't get what the (char) mean
It means "I really do want a char, I promise that I know what I'm doing". Why do you need to tell the compiler that? Because ch is a char and tolower returns an int. Trying to assign a wider type to a narrower type requires a warning, and the cast silences the warning in situations where you really do know what you're doing. Try removing the (char) cast and see what happens, then put it back and compare the results.
And I can't believe no one's told you this yet, but I will however, in lieu of certain "initiatives", shall we say, be nice about it.
When you post source code as if it was regular text, it's very hard to read. The symbols are hard to read, and there is no indentation. Instead, place [/code] at the end of your source code, and [code] at the beginning, and when you post your message, the site will automatically format your code nicely - there's a good example in Prelude's second-to-last post.
Actually no, you wouldn't. You'd get a compiler warning. At least with GCC you will:Quote:
Originally Posted by sean_mackrory, regarding not #include<ctype.h>
However, it still compiles. And, since no one here pays attention to their warnings anyway...Code:ctyp.c: In function `main':
ctyp.c:5: warning: implicit declaration of function `toupper'
Quzah.
could your problem be the fact your using a %c in your printf statement and then not assigning the variable associated with it! Rule1 broken perhaps??? *wags finger*Quote:
Originally Posted by cogeek
also "int main()" is better than just main() as main returns an int.
try that? (ive not yet so i might edit this post yet! .... i AM tired!)Code:#include <stdio.h>
int main()
{
char c;
printf("Add a character: ");
scanf("%c", &c);
toupper(c);
printf("The character is %c\n", c);
return(0);
}
--CHriS
right well theres an improvement already!
guyattc2-practical_c3-> ./file
Add a character: c
The character is c
guyattc2-practical_c3->
now you see your other problem is youve not actually assigned anything to the operation "toupper(c)"
so heres a revision of the code:
and heres the output:Code:#include <stdio.h>
int main()
{
char input;
char output;
printf("Add a character: ");
scanf("%c", &input);
output = toupper(input);
printf("The character is %c\n", output);
return(0);
}
guyattc2-practical_c3-> gcc -o file file.c
guyattc2-practical_c3-> ./file
Add a character: c
The character is C
guyattc2-practical_c3->
game, set and match! *bows* hehe sorry im in a random m00d today! :D
--CHriS
How many days left are there in that month thing?Quote:
Originally Posted by goosematt
Quzah.
» Roughly, 23 days; 10 hours; 56 minutes [GMT -7] As of right now. :cool:Quote:
Originally Posted by quzah
- Stack Overflow