EOF and -1 to count lines, words and characters entered inside the role of language c. We will be simplified definition of the word, that the word is a sequence of characters that does not contain spaces, tabs \ t and newline

**

* On-line, word and character input is counted. We will be simplified definition of the word, that the word is a sequence of characters that does not contain spaces, tabs \ t and newline

**. For example: "a + b + c", that is a word which consists of five symbols. Another example: "xy abc", as two words, six characters. Ctrl + z input end (Tip: ch = EOF with or ch = - 1 as the loop condition!!).

** Output format requirements: "Lines =% d \ nWords =% d \ nChars =% d \ n"
run the following example:
Hi ~
Good Moring!
The I'm Julie.

Lines=3
Words=5
Chars=28

#include <stdio.h>
int main ( )
{
    char ch = '\0';
    
    //循环接受字符
    int l = 0,w = 0,c = 0;
    while (scanf("%c",&ch) != EOF)
           {
        c++;
        if (ch==' '||ch=='\n') {
            w++;
        }
        
        if (ch=='\n') {
            l++;
        }
    }
    printf("Lines=%d\nWords=%d\nChars=%d\n",l,w,c);
    return 0;
}

Note:
1. The receiving transducer line can not be used chOr 13 '
' can only ch'\ the n-'
2 · *** *** EOF by -1 is a meaning in one is enough, end of file , no additional output in the compiler inside, but the operating system can (as I do not know what currently)