1-11 Exercise 3: Prefix of a string

0 like 0 dislike
11.1k views

A string p is a prefix of a string s if the leading characters of s are identical to p. For example, "abc" is a prefix of "abcde". Develop a function prefix that checks if a C string p is a prefix of another C string s.

p是s的前綴若s開頭的一段字串等於p。例如，"abc"是"abcde"的前綴。寫一個程式 判斷字串p是否字串s的前綴。

Example input 1:

abc
abcde

Example output 1:

abc is a prefix of abcde

 

Example input 2:

bc
abcde

Example output 2:

bc is not a prefix of abcde

 

[Exercise] Coding (C) - asked in Chapter 13: Strings by (12.1k points)
ID: 41439 - Available when: Unlimited - Due to: Unlimited
edited by | 11.1k views
0 0
12345
commented by (-329 points)

27 Answers

0 like 0 dislike 
Hidden content!
#include<stdio.h>
#include<string.h>
char prefix(char string1[],char string2[])
{
    int i,s;
** **  ****    * *    * ** **    **
*  ****  **** **   *    ** *  * ****** ***       * ** *  **
    {
   * *      * * * ****** ****  *** ** **** ** ******* * (i=0; i<=strlen(string2)-1;i++)
**** * ** *  *   **** **  ****  *  *  *  **** *  * ** *
 *** ** ** **  * *  *  **    *   * *  ** *** *  ** *   * **  ****   ***** *  (string2[i]==string1[i])
* **** *  ***** *  *   ****** *  * *** *  **  *  * **  * * *****  ** ** * **** *** 
***  *  * *** *  **  *** * ** * * **  *** *   * *  ***  *   * **   *   ** * * *****   **        ** *  **    *  **
   * *   ** * **             *** * * ** ***  ***** ******    ** **** ** ** ** ** * ** 
   * *   *      ** **  ****  * *  * *   *  *  *  ***   * *
 * *  ** **** ***   *  **   ***     ***   * * *   *** 
** * *    **  **  ** * *   ** * ** *  *  *  **  **    
* **   *  *          *  ** *    *  **  ****  * ** *   * **  * ***  *** *  * *   *  *** * *  is a prefix of %s", string2, string1);
   * *  *  ****** **  *  ****   ***  *****   ** *  *
****   *** ****     * *  ***  * *  *   ****** *  *    * **
    ** * ****** **   * *  *  ******** * * * ***   *  * 
*   ******* ** *  **** ** *   * *  **  *** * **      ***** * ***    *****   ***** * *    *   * is not a prefix of %s", string2, string1);
    ** * *  *     **** **  * **  * *  **** *   * ** * *  * 
    }
******  * ** * *  *  ** *      *   *   ***** * * ** * 
    {
*   **     *** ****   ** ** *  *      * *  *  * * ***     (i=0; i<=strlen(string1)-1;i++)
 ****** **  *** *    ** *     ******   ***  ** * 
**      ***   **** *    *  **       ****  *** ***  *** *** * ** ** * *   *******  ****  (string2[i]==string1[i])
      *  ** *     * *   **** ***  *   *   * * *   *   *  **  *  * *** ***   * *  **   
   *** * * *   * *** * **** * **** ***** ***  * *******   * ***  **  ***    ** * **** ******   * ***  *** * * 
**    ** * *** * * ***** * * * * ** ****  ** ***   * *** * *    **    *  * **  
* * *   * **  *******  * *  *  **  *     * ** ** ** ** 
**** ***  ***** *    * *** **      ** **  * *  **  *  **  
* * **   ***  ***     **  *    ******  * **  *   **  *  **
* **     * ** * ***   * *   * *****   * *   **** * **    ****  **  * ** *  *      ****    **    is a prefix of %s", string1, string2);
  **   *  * *  * *     * * * *** * ****** **** *** * * 
*********    *       ****  * * *   *  *  * ****** **** 
 * *      *    **  * *     *  * *  *** ** * * * *****  * 
*** *******   **** **      ** * ** * ** ** * ** ****   * ** *** **  *    *  * * *****     ** is not a prefix of %s", string1, string2);
 *** *  *    ***  ***  ** *    * * * ** **  **  *  
    }
}
int main (void)
{
    char chaine1[1001], chaine2[1001];
  * * *** **  ** *   * **   **    %s", chaine1,chaine2);
*  ***  * *** *   ****  *** * *    ** 
 *** *   * *******  *    * ** *** 0;
}
answered by (-168 points)
edited by
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>

#include <string.h>

int main()
{
      * *   * * ** * *   lenght1,lenght2,c;
* * *   *   * **  * ** ** b[100];
 ****  ****  *  * * ***  *   a[100];
**  ** ** * ***   * *  *  *** d[100];
** *  * *    **** *   *  *  *  *****  **   * * 
 *  * *   *****   *  **** *            **
 ** *   **** * ** *   ***    * ** *** *  
* ***  * *** * *******      **     ** * 
   *    *    ***** * *  * **     ** * ** 
*     * *  * *** ** **  *  * *
    **   ****     * *  *  *  * *     * *   **     * is a prefix of %s",a,b);
*   ** *** ***** ** * **** 
    ** *  * ** *   ***** ****** *
  *****   *  * ** ** *** ** *
 *  ** * * *** *  **    *   *    **   **  * * * ****   *      * *  is not a prefix of %s",a,b);
*  ****** * *  *  ***** ** * *** 
**  *  * *  *  *******  ***  ***  0;
}
answered by (-140 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include     **    * *   *
#include * * ** ** *** *  *
//#include <ctype.h>
#define n 100
int aa(char A[],char B[])
{
*  * *  * *** ** i;
  * ** ** ****  * **   * 
  {
 * * ** ***  *    *   ** **     ***  *  A[i]!='\0' && A[i]!=B[i]   )
  ** **     ****   ** ******** * *    * **  ******  ***  ** 0;
* ******** *  ***    **     **  * *** *  *  *   *   * **
*  ** **  *** ** ***  *          *** **** *** *** * *     *** * *  *   * * *
  }
* ** * ***   **  1;
}


int main()
{
    **    ***  *  * * * * *** * a[n],b[n],c[n];
****        ** * **  **    ** i;
 * ***  * ** **   * * *** *     **  ****   *  *
* ** *  *   *  * * **  * ****** ** * **  * ** *
** **  **  ** ***  *  **  *  * (aa(a,b)){
 * *   *   **** ****  **   ***   ***   **** ** *    * ** *****
 **  ****** *** *** *  ** *  *  * *    ** *  * *    * **      * **   ***  *    *    ******* **
**    **   **  *     *  *  **  * * *   **  *  * ** *  **  ** ****** ** * **** ******* ** ****  ** *   *   **     * * ***  * *  **
 ** * **** * ****** *   *** *     *  **** ** *** ** **  ** *   ***  * ******* ** *** ** 

    *** * ***       *******  * * **** ****   *** *   ***   * *
***   **  * * * **** **** ** **  *  * * *  *   ** ******  ***  *     is a prefix of %s",c,b);
 *   **    **   *  *****  *
*   ******  *     *  ******** 
** * *  ******** * *** ***    **    * **** **** * * *** ***  is not a prefix of %s",a,b);
   *** ** ** *****     *   *    0;
}
answered by (-32 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include  ** * * * ** *
  *  ** * **   *     *  * ***
int main (){
char  *  **
char * **   
  ***   *       string1:  **   **
 *  ***********     ***  *** * **  
* * *    *  *  string2:  ** *    *
 * *  ***  * *** **  * *  **
 ** **    *  ** *  ** 
*   ***   ** *    ********      *    is ****       of ** * *   ***  ** ****
else
 **     * *****    **    * *    ******* is not prefix of ** *********  *    ***

return 0;
}
answered by (16 points)
0 0
prog.c: In function 'main':
prog.c:7:9: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[100]' [-Wformat=]
 scanf("%s",&string1);
         ^
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include   *  ** *   **
#include    *  *  *  ** 
#include ** *  *   **** 
#include **  ** ***** **
int prefix(char p[],char s[])
{
 **  * * *****   *    ***    *  * **       * 
*    **  * ** * * ** * ****** * * 0;
*    ******  * *** *   * if     * * * *  * **
** **    * *****   * *  *****  ** 1;
   ****  *  * ****   *****  *
 *   *   *  * *****    *  **  *** * 0;
}

int main()
{
 ***  * **  *   **   * p[999],s[999];
 **  ***       ** *   * *** ***  * *   * *
**  * **     * **  *   ** * ******   *   

  *     * * *  *** *  *  ***** 
 **  ** ***     * ** ** *   * **  *** * ** is a prefix of %s",p,s);
    *  * ** **  *    * 
  *** * *   ** **** *** *  *  * * *    **  * is not a prefix of    *   ******

   * *   * * ** **** * *** 0;
}
answered by (-16 points)
edited by
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include ***** **   **  * 
int main()
{
**  *        *   * ***  *   * *   length1,length2,c;
 *        *** **** **     a[100];
*    *   * ** ****   * *  ****   b[100];
*   * * *** *  * *******    ** d[100];
** *   **  ***  * * * **  ***** ***** *** *  **
* * * * *  **    ***      * * ****  * ** *     
**   **  * * * **   *******   *  **  
** *  ******* *** *   ***  ***** ***  *    
if (strcmp(a,d)==0)
   * * ***** *   **
   * ** *   **      *  *   *** ***** * *****  *** *  *  * *****  **** *  ** is a prefix of %s",a,b);
* *  *    ********  * ***     
* ** *   **   *****     * **    
** * ** * **  **  *     **    **
  * * ** * * ** ***   ** *** ** ****  * *    **** *** * *  *   * * is not a prefix of %s",a,b);
     *  * * ***** ** *  * *
 ***    ***    * 0;
}
answered by (-229 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include * *** *  ***    
#include *** * *  * *  **

int prefix (char  ***** * str2[])
{
***   **   **      **     **** length=0,   ** 
**  * ** **  *   
  **     *  * * *** **  *** **  = ** ** *** ** * *  
 * ****** * ** * * *  ***  
******     ******** * *  ** *      *        *** 
**  *    * * *  ***   **** *    ***   * **
*       * ** *  * ***  *****  if ** **      
*    ****     **   **        *      **  *  **        
*  ******* ****  **    * * * *  *   }
    *            * *    **  **  *    *
  ** *    **** *    ***  (count == length)
   *  ****   **  *   *   *  ** **  * * ***** 1;
   **** *   *       * * * **  else * ***** **  0;  *  *    ****   **
  *    * * * **     **  *  *  
}
int main(void)
{
* * ****  **  **   **  * ** *      *  * **  ***
* *   * *   * *  *   * ****
*  ***  * * * *  *    ** * * ** *** ***   * str1, str2);

 *  **  * * **   *   * * *    ********  
**  *  **  *   {
**  * ** * ******* *   * ** ***    *  * *  *   * **** ** ***** * **** ** is a prefix of *  ****** str1, str2);
*******   *  ***   *  * *
 ** *      **   *   *   **  ** *
 *   **  ** * *   ***     *** ******** * *  ******   ** ** * * *   * is not a prefix of ** **  * * str1, str2);

*        *  ** ***    **    0;
}
answered by (-304 points)
edited by
0 0
Case 0: Wrong output
Case 1: Correct output
Case 2: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include **    ******  
#include    **  *  ** ***
#include  *  *  *  * *    

int main()
{
**  * *   *  ***  ** ***  ***  * prefix[100];
 **   *   ** *** ** **** **   * word[100];
  ***   * ***   ** ******* *    ***   *  %s", prefix, word);
* * ** ** * *   *  **** * *   * i;
   **   **** ****   ** *   = 0; i< strlen(prefix); ++i)
 ** ** *** * * *  *** ** *  
**      * **    * ****  **  ** * ** *  **** *  ** ** **    *  *
 * *  *  **   * * *  ** ** * ** ** * **   ***  *** ** ***
  **    **      ** ***** *** * ****   ***  *  **** **   *    *    * *  *   ****    * ** *  **   is not a prefix of %s", prefix, word);
* * **    *      * **  ****** * *  **     * ****    ** *  * * * ***   ***   **** ** ** *    0;
 *   *** * * ** *******    ** * *   **  * **  ** *    * * ** * *
*** *       * * *** **  *** 
***  * * ***  ****   **    *         * is a prefix of %s", prefix, word);
**  *  ** ** *  ****  ***   * *** 0;
}
answered by (-108 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include  *    *   ***      
#include **  ***    * *** 
#include ***********  

int main()
{
 ***     *** * *  ***     *  * prefix[100];
 * ** *** ** **   * * ****  word[100];
*    *    * *** *** *** **** *  ** *  **   %s", prefix, word);
    * ***  * * * *   **       i;
** * **        * *** * ** *  *** **  = 0; i< strlen(prefix); ++i)
  *  ***      ******* **** *
* ** *****   ***    * * ** * * **  ** ****** *    ** *  * ***** * *****
 * * ** ***  ******  * *** ** * ********** ***     *** 
** **** *  *** * **  * ** * ** * * ***  ** ** ****  **      **   ** *****   *  * **   ***** is not a prefix of %s", prefix, word);
*** **  * ***   *  **   **  * **** ** * *  *  *   * **    *  ****** ** *  **   *****    0;
  * ** *   **  **  **   **   ***   ** *   ****   * 
* *******  * * ** **  ***
****   *  *  **  **  ** *****  *    is a prefix of %s", prefix, word);
 *  ***   * * *** *  ** *      *  0;
}
answered by (-258 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
 ****     *   *** ***
** ****     ** ****  * * 
**  * *  * *  **    *  

int main()
{
** * **  * **      char   *     
***** * **  char    *  
**  **** ** **   **  * *   *** * **** *     * word);
  ****   * ** int i;
     *    **** for(i = 0; ***  * **  **** * ++i)
 *    *    ***** {
  *** *  **** * ******  *** * * *    * ******* 
  ******   * * *   **       ** * {
 ** **   **   *    **    *  *  * * **** **  *  ******* is not a  *  of ****   **    ***  word);
   ** **   **   **  **   *      * **  *  ***    ** * 0;
*   ** ** * *  *       *   }
** * ***   ***  * }
    **  * *** *   *** ****  ** is a prefix of      * *   *  word);
* *  *    ****** *  ** 0;
}
answered by (-249 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
commented by (4.3k points)
Page:
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:108.162.241.33
©2016-2025

Related questions

0 like 0 dislike
57 answers
1-11 Exercise 1: UPPER and lower letters
[Exercise] Coding (C) - asked Jan 11, 2018 in Chapter 13: Strings
ID: 41431 - Available when: Unlimited - Due to: Unlimited | 16.2k views
0 like 0 dislike
29 answers
1-11 Exercise 2: Reverse a sentence
[Exercise] Coding (C) - asked Jan 11, 2018 in Chapter 13: Strings by thopd (12.1k points)
ID: 41436 - Available when: Unlimited - Due to: Unlimited | 11k views
0 like 0 dislike
53 answers
1-4 Exercise 3: String details
[Exercise] Coding (C) - asked Jan 4, 2018 in Chapter 13: Strings by semicolon (5.2k points)
ID: 40715 - Available when: 2018-01-04 18:00 - Due to: Unlimited | 14.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users