Prog to list ASCII codes (beginners' stuff)

Question

Anu 0 Newbie Poster

19 Years Ago

Hi,

I just started a program containing every single prog we have done for school so far and added some private stuff.
Searching through my older files I found a prog which lists all Integers between 1 and 255 and their respective ASCII char, so I decided to include it into my 'collection'.

The problem is that 'twas a Win32 executable and we're working under DOS. Well actually this wouldn't be much of a problem, but the script seems no to work under DOS.

Well, first of here's the script:

# include <iostream.h>
   # include <conio.h>

   main()
   {
      int i, j, ip, ipp;
      float k, l;
      char zeichen;
      for (i=1;i<=254;i++)
      {
          zeichen=i;
          cout<<"\n"<<i<<" - "<<zeichen;
          j=1;
          for (j=1;j<=25;j++)
          {
               k=1.0*i/20;
               l=1.0*j;
               if (k==l)
               {
                    ip=i+1;
                    ipp=i+20;
                    if (i<=240) cout<<"\n\n(Press any key -->  "<<ip<<" to "<<ipp<<" )";
                    getch();
                    clrscr();
               };
           }
        }
        getch();
   }

As I stated, no problems under Win32, but under DOS it stops at 26 and after pressing a key again the screen gets cleared again and stays black.

So my question is..is there any way to get something like this running under DOS?

I tried a simple "cin>>intvar; charvar=intvar; cout<<charvar;" aswell, but any value above 25 simply results in nothing being put out.

Well, thanks in advance.~

c

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Latest Post 19 Years Ago Latest Post by chound

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Dave Sinkula 2,398 long time no c

19 Years Ago

Skip 26. Or better yet, skip printing everything non-printable.

#include <stdio.h>
 #include <ctype.h>
 
 int main(void)
 {
    int i;
    for ( i = 0; i < 256; ++i )
    {
 	  putchar(isprint(i) ? i : '.');
 	  putchar(i % 16 == 15 ? '\n' : ' ');
    }
    return 0;
 }
 
 /* my output
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
   ! " # $ % & ' ( ) * + , - . /
 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 { | } ~ .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 . . . . . . . . . . . . . . . .
 */

Dave Sinkula 2,398 long time no c

19 Years Ago

[Oops. I see your reply beat my edit.]

I'd use isprint.

#include <stdio.h>
 #include <ctype.h>
 
 int main(void)
 {
    int i;
    for ( i = 0; i < 128; ++i )
    {
 	  putchar(isprint(i) ? i : '.');
 	  putchar(i % 32 == 31 ? '\n' : ' ');
    }
    return 0;
 }
 
 /* my output
 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
   ! " # $ % & ' ( ) * + , - . / 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 { | } ~ .
 */

Stack Overflow 8 Junior Poster

19 Years Ago

Check out http://www.asciitable.com/

The full list of ASCII characters are there. The function isprint() knows which of the characters are printable. isprint() returns a nonzero value if the input is a printable character, including the space character (0x20 – 0x7E). The other characters are either extended, or do other features which aren't print friendly.

More information on isprint()
isprint() is included in the ctype library, ctype.h:

int isprint(int c);
» This function returns nonzero if c is a space or a character for which either isalnum() or ispunct() returns nonzero.

int isalnum(int c);
» This function returns nonzero if c is any of or other locale-specific alphabetic character.

[i]According to AsciiTable.com[/i]
[u]Characters[/u]	[u]ASCII[/u]
a thru z	97 thru 122
A thru Z	65 thru 90
0 thru 9	48 thru 57

int ispunct(int c);
» This function returns nonzero if c is any of the following or other implementation-defined punctuation character:

[i]According to AsciiTable.com[/i]
[u]Characters[/u]			[u]ASCII[/u]
! " # $ % & ' ( ) * + , - . /	33 thru 47
: ; < = > ? @			58 thru 64
[ \ ] ^ _ '			91 thru 96
{ | } ~				123 thru 126

Hope this helps,
- Stack Overflow

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Anu 0 Newbie Poster · Answer 1 · 2004-09-25T03:44:02+00:00

Skip 26. Or better yet, skip printing everything non-printable.

Wow, thanks for the fast reply.

Well, but how to skip them most easily?

The first and more or less only thing that comes into my mind is like:

if ( i != 26 && i != anotherone && i != yetanotherone &&...

Anu 0 Newbie Poster · Answer 2 · 2004-09-25T04:32:54+00:00

I see, thanks again.

But seeing this...

The output under Win32 is like:

1   - ☺ 
2   - ☻ 
3   - ♥ 
4   - ♦
5   - ♣
6   - ♠
7   -
8   -
9   -
10  -
11  - ♂
12  - ♀
13  - 
14  - ♫
15  - ☼ 
16  - ► 
17  - ◄ 
18  - ↕
19  - ‼
20  - ¶
21  - §

And so on, so...

Aint I able to display the 255 ones of those under Dos?

Anu 0 Newbie Poster · Answer 3 · 2004-09-25T05:40:52+00:00

Thanks, might be quite of use, though probably not directly in what I'm thinking of.

See the attachments, I just want the fourth attachment (Dos, page2) to be like the second one (W32, page2).

Dave Sinkula 2,398 long time no c Team Colleague · Answer 4 · 2004-09-25T06:25:49+00:00

Check out http://www.asciitable.com/
The full list of ASCII characters are there. The function isprint() knows which of the characters are printable. isprint() returns a nonzero value if the input is a printable character, including the space character (0x20 – 0x7E). The other characters are either extended, or do other features which aren't print friendly.
More information on isprint()
isprint() is included in the ctype library, ctype.h:
int isprint(int c);
» This function returns nonzero if c is a space or a character for which either isalnum() or ispunct() returns nonzero.
int isalnum(int c);
» This function returns nonzero if c is any of or other locale-specific alphabetic character.
[i]According to AsciiTable.com[/i]
 [u]Characters[/u]	[u]ASCII[/u]
 a thru z	97 thru 122
 A thru Z	65 thru 90
 0 thru 9	48 thru 57
int ispunct(int c);
» This function returns nonzero if c is any of the following or other implementation-defined punctuation character:
[i]According to AsciiTable.com[/i]
 [u]Characters[/u]			[u]ASCII[/u]
 ! " # $ % & ' ( ) * + , - . /	33 thru 47
 : ; < = > ? @			58 thru 64
 [ \ ] ^ _ '			91 thru 96
 { | } ~				123 thru 126
Hope this helps,
- Stack Overflow

The greater utility of functions like ispunct and isalnum are that they also work with other character sets, some of which, for example, do not necessarily have a contiguous alphabet or such. They are designed to work regardless of the particular implementation.

Dave Sinkula 2,398 long time no c Team Colleague · Answer 5 · 2004-09-25T06:28:43+00:00

See the attachments, I just want the fourth attachment (Dos, page2) to be like the second one (W32, page2).

DOS has a particular issue with the value 26. Apparently it is an end-of-file marker. So on a DOS system, it will not work the way you'd like. Thus my recommendation to skip it.

Anu 0 Newbie Poster · Answer 6 · 2004-09-25T16:15:22+00:00

That was all I actually wanted to know. Thanks once again. ^^

Anu 0 Newbie Poster · Answer 7 · 2004-09-25T20:13:01+00:00

Just wanted to say, I just skipped all non-printables using an int array and it works just as it's supposed to.

Might be getting annoying by the time, saying it in every post, but thanks again. Nice board you got here, I -as a very beginner- learned a lot of useful stuff by just reading some threads. -which also saved you e.g. another "void main <--> int main ?" thread- :mrgreen:

chound 1 Junior Poster · Answer 8 · 2004-09-26T08:25:04+00:00

You could use stdio.h and
for(int i=0;i<253;i++)
{
printf("%c", i);
}

Prog to list ASCII codes (beginners' stuff)

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