Printing only certain characters

Hello Everyone,

Is there a command that will print only certain characters. I am looking at the strstr command but not sure if this is right.

What I am trying to do is have the user enter a string of characters and then print only ones that end in 'ed'.

I also looked at tokenized and then use strstr to parse out only strings that end in 'ed'. Hopefully I am on the right track.

An advice would be appreciated

Thanks

DMKanz07

Sounds like a good plan to me.

Okay I am stuck on how i can read a series of strings and print only those strings that ended with the letters "ed"

Here is what I thought I could do is find the length of the string and the validate the last two characters. But I am not sure how to do that.

Oh C gods show me the way

Code:

---

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

/* Function main begins program execution */
int main()
{

    char s [ 100 ];
    int i;
    printf( "Enter a line of text: " );

    /* Use gets to read line of text */
    gets( s );
    printf( "\n" );
   
    for( i = 0; i < strlen ( s ); ++i ) {
        printf( "%c\n", s [ i ] );
    }
   
    printf( "%lu\n", strlen( s ) );
     
    /* Indicates successful termination */
    return 0;

} /* End Main */

---

Thanks

DMKanz07

Ok. Again, don't use gets(). It's bad news.

But for what you want to do, use strlen to find the length of the string and suppose you store that in len.

After that, you know that s[len - 1] will be the place where there is a null character.

Therefore, s[len - 2] is the last character and s[len-3] is the second to last. Done.

heres an example

Code:

---

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

int main()
{
    char s[10];
    char last, secondLast;
    int len;
   
    fgets(s,sizeof(s),stdin);   
   
    len = strlen(s);

    if (len < 3)
    {
          printf("enter word with at least 2 characters\n");
          return 1;
    }
   
    last = s[len - 2];
    secondLast = s[len - 3];
 
    printf("last char is: &#37;c\n", last);
    printf("secondLast char is: %c\n", secondLast);
 
    // maybe use some if statements to compare these last two characters?
}

---

Okay I have been trying to figure out how you test to see if the last to character are equal to 'ed' and I am stumped

I tired

Code:

---


if ( secondLast = "e" && last = "d" )
    printf( "%c\n", s [ i ] );

---

but it does not compile. I am not understanding this very well. Sorry guys.

DMKanz07

when you say 'it does not compile' always remember to post why it doesnt compile! ie, what does the compiler complain about?

in this case, secondLast and last are of type 'char'. when you assign values to chars or check their values, use single quotes.

Code:

---

char myChar = 'a';

---

hope it helps

> if ( secondLast = "e" && last = "d" )
1. Use == for comparing, not =
2. A single character uses single quotes ('d'), not double quotes ("d")

Also, I thought you were going to work in 2 stages
- get a list of all the words
- print those which end in "ed"

Code:

---

if ( secondLast = "e" && last = "d" )
    printf( "%c\n", s [ i ] );

---

I don't know why there's an i in there. You don't have to loop through all the characters in the string. The last and second last are always given by what I've told you beforehand. Always.

Nadroj,

I am but I just trying to get one word to work first. Dang I know == compares dont know why I used =.

C is still very new to me, i should say programming is new to me

Thanks for your patience

DMKanz07