An English statement calculates the number of words in the input

1  // 
2  //   main.c
 3  //   count the number of words entered
 4  // 
5  //   the Created by Longma ON 2019/6/27.
 6  //   Copyright © 2019. All Rights Reserved.
 7  //
 8  
9 #include <stdio.h>
 10 #include < String .h>
 . 11  
12 is  int main ( int argc, const  char * the argv []) {
 13 is      
14      // input a sentence in English, English sentences separated by a space. Count how many words in a sentence.
15      // EG: How are you, there are three words 
16      char String [ 80 ];
 . 17      the printf ( " Please enter the English sentence: " );
 18 is      the gets ( String );
 . 19      
20 is      char C;
 21 is      char kongGe [ 2 ] = "  " ;
 22 is      int Word = 0 ;
 23 is      int NUM = 0 ;
 24      
25      for ( int I = 0 ; (C = String [I]) =! ' \ 0 ' ; I ++ ) {
 26 is         IF (C == 32 ) { // the current bit spaces (ASCII code value of 32), word recorded as 0. Note: xcode in strcmp (& c, "") - 120, is not 0, strcmp comparison It is a string, and the string of characters can not compare! 
27              the printf ( " % D,% D equality:% D \ n- " , C, kongGe [ 0 ], strcmp (& kongGe [ 0 ], & C));
 28              Word = 0 ;
 29          } the else { // current bit It is not a space 
30              IF ( 0 == Word) { // the space of a / the first letter is 
31 is                  NUM = + . 1 ;
 32                  Word = . 1;
 33 is              } the else { // the space is not one, can not be considered a word
 34 is                  // do Nothing 
35              }
 36          }
 37 [      }
 38 is      
39      the printf ( " statement:% S \ n- " , String );
 40      the printf ( " number of words:% d a " , NUM);
 41 is      return  0 ;
 42 is }

log:

Please enter English sentence: warning: the this Program uses the gets (), Which IS unsafe. 
How are you 
32, 32 are equal: -120 
32,32 equality: -120 
statement: how are you 
word count: 3 Program ended with exit code: 0

 

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Origin www.cnblogs.com/Dast1/p/11100514.html