1-11 Exercise 3: Prefix of a string

0 like 0 dislike
10.9k 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 | 10.9k 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:172.69.7.77
©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 | 15.8k 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 | 10.7k 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 | 14.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users