Thread: Character program

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

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;
    }