1-11 Exercise 3: Prefix of a string

0 like 0 dislike
12.9k 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 | 12.9k 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:104.23.243.108
©2016-2026

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 | 18.7k 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 | 12.4k 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 | 16.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users