使用IN / OUT标志每行显示一个单词一个字符串

Question

I have written a program that first stores an arbitrary number of lines of text from the user. After that, it checks when a new word has come and if it does, then it prints it in a new line. Below is my code:

 #include<stdio.h>
 #define IN 1   //inside a word
 #define OUT 0  //outside a word

int main()
{
  char s[100];   //storing the string entered by user
  int c;         //getchar reading variable
  int i=0;       //iterating variable

  while((c=getchar())!=EOF)
  {
    s[i++]=c;
  }
  s[i]='\0';   //end of string


  i=0;
  int p=0;    //stores the start of the word
  int current=OUT; //flag to indicate if program is inside or outside a word

  while(s[i]!='\0')  
  {
    if(current==OUT)  //when program is outside a word
    {
        if(s[i]!=' ' || s[i]!='\n' || s[i]!='\t')  //word found
        {
            p=i;  //store starting position of word 
            current=IN;
        }
    }
    else if(current==IN) //program is inside a word
    {
        if(s[i]==' ' || s[i]=='\n' || s[i]=='\t') //end of word found
        {
            current=OUT; //flag now outside the word
            for(int j=p;j<i;j++) //print from starting position of word
            {
                printf("%c",s[j]);
            }

            printf("\n");  //go next line

        }

    }
    ++i;  //incremnent the iterator variable
 }

return 0;
}

My program works well if I just enter the string in a proper manner, ie without any extra spaces or new lines. But if I enter a line as follows ( notice the extra spaces and new lines ):

*I am a boy

I went to Japan */

Then it prints those extra newlines and spaces along with word too, which according to me should not happen because of the IN and OUT flags. The output is like this: enter image description here I request you to please help me out.

I know I can do this easily with the putchar() method checking one character at one time, but I am just curious as to what I am doing wrong in this implementation.

Answer 1

First bug that jumps out at me:

if(s[i]!=' ' || s[i]!='\n' || s[i]!='\t')

will always return true. You want && , or else use a !() around the whole condition that you use the other place, for symmetry.

Or better yet, factor that out into a function, or use isspace from <ctype.h> .

Answer 2

Your filtering condition for determining if a character is white space is not correct. The || operator means OR. Using chained OR will allow the expression to evaluate to true every time. You need the AND operator && . The and operator fails as soon as one operand evaluates to false, or in the case of C, 0 .

Besides that, there are better ways to check for white space. One idea is using the isspace function from <ctype.h> , which accepts a character as an int , which can also be an unsigned char , and determines if that character is any of ' ', '\\t', '\\v', '\\n' or '\\r' . You can also do character checking via switch statements

switch(ch) {
    case ' ':
       // do something
        break;
    case '\n':
      //do something
        break;
}