Character program

Hi there,
I have to make c program that asks to enter a character and find out is it a number, letter or special character.:)

Here is what i tried:

Code:

---

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

main(void)
{
  char ch,a,z,c;

  printf("Enter a character\n");
  scanf("%c", &ch);

  if (c >= 'a' && c <= 'z'){
    printf("%c is a letter.\n", ch);
  }
  if (c >= '0' && c <= '9'){
    printf("%c is a number.\n", ch);
  }
    switch(ch)
  {
    case '"':
    case '{':
    case '(':
    case ')':
    case '&':
    case '^':
    case '%':
    case '*':
    case '#':
    case '!':

      printf("%c is a special character.\n", ch);
      break;
  }

    return 0;
}

---

Well for one you should have break; inside of each case.

Do you have a specific question?
And would you be willing to try some specialized functions for this purpose?
isdigit - C++ Reference
isalpha - C++ Reference
ispunct - C++ Reference

No, just this example above the task. Attachment 12299

I still dont understand what your question is?

You need to be more specific.

These if loops doesnt work. Is there any other mistakes?

"if()" is a conditional statement, not a loop.

Line 9: You read a character into a variable called 'ch'
Line 11: You use 'c' (not 'ch') in your "if()" statement
Line 14: You use 'c' (not 'ch') in your "if()" statement

Quote:

---

Originally Posted by hbranum

Well for one you should have break; inside of each case.

---

This is not true.

Not every case in a "switch()" statement should have a "break." It depends on how you want it to behave. The OP is exploiting the "fall-through" approach of a "switch()" statement.

Sometimes, you want one thing to happen for several different values.

Here's a quick example (it's poor code, but illustrates the concept):

Code:

---

char input;
int exit;

printf("Continue?  [Y/N] -> ");
scanf("%c",&input);

switch(input)
{
    case 'Y':
    case 'y':
        exit = 0;
        break;
    case 'N':
    case 'n':
        exit = 1;
        break;
    default:
        break;
}

---

"exit" will be set to zero if the user enters 'Y' or 'y'.
"exit" will be set to one if the user enters 'N' or 'n'.

I didn't understand whats wrong in line 9, but seems like it works anyway:)

Code:

---

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

main(void)
{
  char ch,a,z,c;

  printf("Enter a character\n");
  scanf("%c", &ch);

  if (ch >= 'a' && ch <= 'z'){
    printf("%c is a letter.\n", ch);
  }
  if (ch >= '0' && ch <= '9'){
    printf("%c is a number.\n", ch);
  }
    switch(ch)
  {
    case '"':
    case '{':
    case '(':
    case ')':
    case '&':
    case '^':
    case '%':
    case '*':
    case '#':
    case '!':

      printf("%c is a special character.\n", ch);
      break;
  }

    return 0;
}

---

Nothing was wrong with line 9 - I just wanted you to see that the variable you were reading into was not the same variable you were using for comparison.

oh yes,Thank you. So is this code ok now? There is no other option for "special characters"? I have to write one by another in every case " "?

Quote:

---

Originally Posted by Solarwin

So is this code ok now?

---

It appears to be okay. I'm assuming you tested it. I'm also assuming the character variables 'a', 'z', and 'c' are there because you plan on growing this program (they have no use in the program thus far).
Some sticklers might recommend a "default" case in your "switch()" statement for consistency, even if it just executes a "break", but for purposes of this code, it's alright without it.

Quote:

---

Originally Posted by Solarwin

There is no other option for "special characters"?

---

You don't have specifications/instructions to tell you this? There are others you could add ('+' '-' '$' '.' for instance)

Quote:

---

Originally Posted by Solarwin

I have to write one by another in every case " "?

---

I don't understand this question. Can you clarify?

Nope, i wasn't planning on growing this.:)
I dont have any other instructions.I think i have to add all of C Character Set these special characters. I was wondering if there is any way how to add them without using "switch" and "cases". It is kind of boring to copy and paste all of these characters.

Code:

---

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

main(void)
{
  char ch;

  printf("Enter a character: ");
  scanf("%c", &ch);

  if (ch >= 'a' && ch <= 'z'){
    printf("%c is a letter.\n", ch);
  }
  if (ch >= '0' && ch <= '9'){
    printf("%c is a number.\n", ch);
  }
    switch(ch)
  {
    case '"':
    case '{':
    case '(':
    case ')':
    case '&':
    case '^':
    case '%':
    case '*':
    case '#':
    case '!':

      printf("%c is a special character.\n", ch);
      break;
      default:
      break;

  }

---

Programming is exciting, but it can be tedious at times. Sometimes, you have to accept that. You do have some options, though.

1. Copy and paste them all. Time-consuming, but easy and guaranteed to meet the requirements.

2. Look up the values on the ASCII chart, and do more ">= and <=" comparisons to check case values in bulk. This is somewhat more elegant than the "switch()" implementation you have, but it's not guaranteed to be universal, as character sets may vary on some machines.

3. Use the "ispunct()" function I linked in post #3. This would be the easiest and nicest way to check, since it seems to check for all the characters you need it to for the "special" case. There's a link there to another page that has a table which lists which symbols "ispunct()" is valid for.

I choosed third option.:) Try to enter number or letter You will see the mistake then.:/

Code:

---

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

main(void)
{
  int i=0;
  int cx=0;
  char ch;

  printf("Enter a character: ");
  scanf("%c", &ch);

  if (ch >= 'a' && ch <= 'z'){
    printf("%c is a letter.\n", ch);
  }
  if (ch >= '0' && ch <= '9'){
    printf("%c is a number.\n", ch);
  }

  char str[]="\n";
  while (str[i])
  {
    if (ispunct(str[i]))ch++;
    printf("%c is a special number.\n", ch);
    i++;

    return 0;
    }
    }

---

First of all, your "return 0" is in your while loop. Good indentation is the key to better seeing the logical flow of your program and avoiding such mistakes. Here is what your code looks like with proper indentation:

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
 
main(void)
{
    int i=0;
    int cx=0;
    char ch;
 
    printf("Enter a character: ");
    scanf("%c", &ch);
 
    if (ch >= 'a' && ch <= 'z'){
        printf("%c is a letter.\n", ch);
    }
    if (ch >= '0' && ch <= '9'){
        printf("%c is a number.\n", ch);
    }

    char str[]="\n";
    while (str[i])
    {
        if (ispunct(str[i]))ch++;
        printf("%c is a special number.\n", ch);
        i++;
 
        return 0;  /* this should not be inside the loop! */
    }
}

---

Second of all, it looks like you just copied the example from that link, without any idea of what you were doing. That example deals with a string, you're just checking a single character. Don't make the program more complicated than it has to be.

Instead of that whole "while()" loop mess, you can solve this with just two lines of code.

Slow down, take a deep breath, and look carefully at how the function is defined.

>> int ispunct ( int c );
>> 'c' - Character to be checked
>> Returns a value different from zero (i.e., true) if c is a punctuation character. Returns zero (i.e., false) otherwise.

So you need to figure out how to test the character with the "ispunct()" function. If the "ispunct()" function indicates that the character is indeed punctuation, print the appropriate message.

Code:

---

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

main(void)
{
  int i=0;
  int cx=0;
  char ch;

  printf("Enter a character: ");
  scanf("%c", &ch);

  if (ch >= 'a' && ch <= 'z'){
    printf("%c is a letter.\n", ch);
  }
  if (ch >= '0' && ch <= '9'){
    printf("%c is a number.\n", ch);
  }

  char str[]="";

  {
    if (ispunct ( int ch ));
    printf("%c is a special number.\n", ch);
    i++;


    }
    return 0;
    }

---

Nothing works :( Can u write that ispunct() fuction?:(

Oh dear. Yes I can, but I don't want to show you disrespect by taking away a good learning experience. But here's a huge clue, because I'm not going to be available for the rest of the night.

Code:

---

if (/* something */){
    printf("%c is a special character.\n", ch);
  }

---

You just have to figure out the "something" part.

And FYI, there are to big problems with this line:

Code:

---

if (ispunct ( int ch ));

---

1. When you pass an argument to a function, you just need to indicate the value/variable, not the data type (i.e. remove the "int").
2. You have a semi-colon at the end of this statement. While the argument to the "if()" statement will be evaluated, even if it's true, nothing will happen. It's equivalent to this:

Code:

---

if (TRUE)
    ;  // <--- empty statement
next line of code  // <--- will get executed because it's not part of the "if()" statement body

---

There are a few excellent ways to reduce your boredom. :cool:

First one is to include <ctype.h>, which has all kinds of functions for testing chars: isdigit(), ispunct(), isalpha(), isupper(), islower(), isctrl(), etc.

And all the above, (and more), can be combined with && , etc.

Now for the special chars, one way to deal with them easily, is to put them all into a single char array:

Code:

---

char special[]={"!@#$%^&*()"}; //put em all in there

---

#include (string.h> in your program

then

Code:

---

len=strlen(s);

for(i=0;i<len;i++) {
  if(strchr(special, s[i])) {
      printf("Yes, it's a special char\n");
  else
      printf("No, it's not a special char\n");
}

---

Another way, mentioned by Matticus above, is to use the ASCII values. Easy to use them indirectly.

Say I want only digits from a string:

Code:

---

int i, len;
char s[]={"abcd12345efghijkl98pqrstuv54321wxyz");

len = strlen(s);

for(i=0;i<len;i++) {
  if(s[i] >= '0' &&  s[i] <= '9')
      printf("Yep! %d is a digit. \n", s[i]);
}

---

When you get into a situation where you are repeating tedious code, it's best to stop, and re-consider. Because the guys who created C were VERY bright, and they didn't want to do a bunch of tedious repetitive code, any more than you do (probably a lot less).

Chances are, they made a function to do what you need. Not always, but very likely.

To add something, a small detail is that your program doesn't seem to handle all types of letters ......upper/lower.

I liked this way: Can somebody find my mistakes?

Code:

---

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

main(void)
{
  int i, len, s;
  int cx=0;
  char special[]={"!@#$%^&*()"}; //put em all in there
  char ch;

  printf("Enter a character: ");
  scanf("%c", &ch);


char s[]={"abcd12345efghijkl98pqrstuv54321wxyz()"};

len = strlen(s);

for(i=0;i<len;i++) {
  if(s[i] >= '0' &&  s[i] <= '9')
      printf("Yep! %d is a digit. \n", s[i]);
}

    return 0;
    }

---

Well s can be one thing, not two. You have declare it as an int AND as an array of chars. This can not be done.