Thread: While loop

Simply put, this exercise demands that only a single while loop is used. The ASCII table is larger than 20 characters, therefore you need modulo to know when to break to the next line. I would do it at (i+1) % 20 == 0, but that's not the only way.

Originally Posted by Milo7

I see how you are using modulo to get a value up until 20, although how would you use 'printf' to print 20 characters and only after 20 characters instruct it to go to the next line using \n ?

I see you are confused so let me just say that you don't have to add a newline every time you use printf. When you don't, the next character will still get printed on the same line as the last. For example, the following code will print all (printable) ascii characters on the same line. What do you need to add to make it print only 20 characters per line?

Code:

#include <stdio.h>

int main(void)
{
    int i;
    for (i = 0; i < 94; i++) {
        // Don't mind the way I print the characters here if you can't understand it, did it this way to make it easier adding your solution
        putchar(i+33); // Same as 'printf("%c", i+33);'
    }

    return 0;
}

Originally Posted by Milo7

this would be the final code.

Code:

#include <stdio.h>
#include <stdlib.h>


int main()
{
  int  count = 33;
while(count<255){


printf("%c ",count);


count++;
if ((count-33) % 20 == 0){


printf("\n");


}
//count++;


}
return 0;
}

what are you printing out again? number off of your count or ASCII table?

%c prints out what data type, or what is printf looking for when using %c
and how is 'int count' the ASCII table?
that count looks funny too.

you're mixing between me and GReaper . where GReaper did the same thing but used a different method. where neither did the complete code for you. only showed you examples to keep and/or check count to get a new line, and that was it.

I think, you need to learn the basics a little more before trying hairball stuff like that, But, sometimes trying hairball stuff like that helps to learn the basics better, so it's give and take,or live and learn.

OOPs, I stand corrected: I just look up how to print out ascii

The %c is the format string for a single character, and %d for a digit/integer. By casting the char to an integer, you'll get the ascii value. To print all the ascii values from 0 to 255 using while loop. Chars within single quote ('XXXXXX'), when printed as decimal should output its ASCII value.

so I'd be something like this;

Code:

int count = 0, check = 0;
char something;
while ( count < 255)
{

printf("%d ", (int)something);

if ( check == 20)
{
     printf("\n");
    check = 0;
}
check++;
count++;
}

BUT : BIG BUT, that is NOT using the criteria for the assignment, which is using modular division

and you need to make that something chars within the ASCII table so it can print out its numeric values.

I think Milo7 nailed it( ignoring their non-existent indentation ). The solution they implemented works just fine, although I would go for "count < 127" because official ascii values are from 0 to 128, not 256.

Originally Posted by GReaper

I think Milo7 nailed it( ignoring their non-existent indentation ). The solution they implemented works just fine, although I would go for "count < 127" because official ascii values are from 0 to 128, not 256.

his code gets this:

Code:

userx@~/bin <> ./a.out
! " # $ % & ' ( ) * + , - . / 0 1 2 3 4 
5 6 7 8 9 : ; < = > ? @ A B C D E F G H 
I J K L M N O P Q R S T U V W X Y Z [ \ 
] ^ _ ` a b c d e f g h i j k l m n o p 
q r s t u v w x y z { | } ~  € � ‚ ƒ „ 
… † ‡ ˆ ‰ Š ‹ Œ � Ž � � ‘ ’ “ ” • – — ˜ 
™ š › œ � ž Ÿ   ¡ ¢ £ ¤ ¥ ¦ § ¨ © ª « ¬ 
.. ® ¯ ° ± ² ³ ´ µ ¶ · ¸ ¹ º » ¼ ½ ¾ ¿ À 
Á Â Ã Ä Å Æ Ç È É Ê Ë Ì Í Î Ï Ð Ñ Ò Ó Ô 
Õ Ö × Ø Ù Ú Û Ü Ý Þ ß à á â ã ä å æ ç è 
é ê ë ì í î ï ð ñ ò ó ô õ ö ÷ ø ù ú û ü 
ý þ userx@~/bin <>

as I thought it was the other way around, using the alphabet to get the numeric values printed out not using he numeric values to get the chars printed out.
where this

Code:

int c = 0;
 char ascc [6] = { 'a' , 'A', 'b', 'B', 'c', 'C'};

while ( c < 6)
{
    printf("%d ", (int)ascc[c]);
    c++;
}

prints this:

Code:

97 65 98 66 99 67

lower 'a' is 97
Cap 'A' is 65

table reference
Reference: Nonprintable and Printable ASCII Characters - Technical Documentation - Support - Juniper Networks

Originally Posted by christop

ASCII is defined for characters 0 through 127. ASCII is a 7-bit standard, so no characters exist outside of that range. Anything or anyone that claims otherwise is uniformed.

How characters outside of that range are displayed depend on the character encoding setting of your system. My system is configured as UTF-8, and each byte value above 127 is printed as a question mark inside a box. Your system appears to be configured to use the ISO-8859-1 character encoding or a similar encoding. That is an 8-bit character encoding that defines codes 128 through 255. The lower 128 characters are identical to ASCII (same as UTF-8), so it's "backward compatible" with ASCII.

so that is why I get them little '?' sometimes, and yes I just reinstalled (Linux) and have not changed it to UTF-8 (yet) ...it is still set to just en_US. I might as well do it right now, Now that you bring it up. because I have ran into a few probs and had to have it UTF-8 ... so thanks for that sharing.