print words of a string

**xiromdim** · 12-04-2013

I want to print the words of a sentence, given as a string....
But I have a problem with the end of the sentence, and cannot find the bug....
Thanks in advance...

Code:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
char str[80]="This is a sentence";
int main(){
    
    char *temp;
    char *word;
    int i;
    
    word=str;
    temp=word;
    while(temp != NULL){
      while(isalpha(*temp)){
           temp++;
      }//we are in a space character
      i=0;
      //print the word
      while((word+i)!=temp){
         printf("%c",(*(word+i)));getchar();
         i++;
      }
      putchar(*temp);getchar();
      while(isspace(*temp)&&temp!=NULL) temp++;//search for the next alphanumeric character
      
      word=temp;
      printf("*temp=%c %d\n",*temp,*temp);
     }
    getchar();getchar();
    
    return 0;
}

**Matticus** · 12-04-2013

The pointer "temp" does not magically have the value NULL at the end of the string.

Instead of checking if "temp" is NULL, I'd recommend checking if the value pointed to by "temp" is a null character ('\0'), signifying the end of the string.

Also, your indentation is good, but I'd suggest using more vertical whitespace. There's no reason to cram two commands on one line (such as on line 20).

[edit] Oh, and there's no reason to have your character array global. That could, and should, be in "main()".

**xiromdim** · 12-04-2013

Thanks very much the answer....
I fixed it...
But what is the "magic" reason that temp does not have value NULL at the end of the loop?
You have right for the identation, as for the global function....
I thought that if str was global temp==NULL will work (lol!!)

**tabstop** · 12-04-2013

The only way you are changing temp is to move it forward through the string. If you start somewhere and move forward, you can never reach nowhere (which is what NULL is) -- it would require "magic" for that to happen.

**xiromdim** · 12-04-2013

I thought that *p equals '\0' is equivalent to p points to NULL.
That means, that '\0' is the NULL character, but NOT the NULL pointer (as a pointer).

**c99tutorial** · 12-04-2013

You might have an easier time if you rewrite the loop as follows:

Code:

for (temp = str; *temp != '\0'; temp++) {
...
}

Then, instead of doing "temp++" inside your loop body, do "continue" instead. This way, it's easy to see that each iteration through the loop has a valid char *temp to work with.

Yes, '\0' has the same integer value as NULL, but notice the comparison:

Code:

*temp == NULL

It does not match in type, because the left-hand side is a char, but the right hand side is a void*

Code:

*temp == '\0'

This way, although equivalent, is better because the types match. Matching types is a good thing in a statically typed language. It helps ensure correctness, for this reason people usually say to turn on all the warnings and never ignore them. Maybe it's a pain at first, but in the end you'll thank the warnings.

**xiromdim** · 12-04-2013

I see..
thanks!

**laserlight** · 12-04-2013

Originally Posted by xiromdim

I thought that *p equals '\0' is equivalent to p points to NULL.

No, it isn't: if *p == '\0', then p points to something, hence it is not a null pointer. If p is a null pointer, i.e., p == NULL (which is not really the same as saying that "p points to NULL"), then *p would be wrong since a null pointer cannot be validly dereferenced.

Originally Posted by xiromdim

That means, that '\0' is the NULL character, but NOT the NULL pointer (as a pointer).

Yes.

Originally Posted by c99tutorial

Then, instead of doing "temp++" inside your loop body, do "continue" instead. This way, it's easy to see that each iteration through the loop has a valid char *temp to work with.

This should work, but I suggest that you use continue as a last resort: usually it is possible to just structure your logic such that the rest of the loop body is implicitly skipped without the use of continue, e.g., by negating the condition and having the if statement body contain the part to be skipped.

Originally Posted by c99tutorial

It does not match in type, because the left-hand side is a char, but the right hand side is a void*

Technically, the right hand side might be an int, but it is still a null pointer constant, hence the types do not match from a conceptual point of view.

Originally Posted by c99tutorial

This way, although equivalent, is better because the types match.

Technically, the types are not the same: the left hand side is a char, the right hand side is an int. However, conceptually '\0' is a character constant, even though its type is actually int, so in that sense the types match.

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