12-21 Exercise 1: Finish the code to match the input and output. Pointer is required.

Question

18.1k views

We will use this function to proceed and store the result

void find_two_largest(int a[], int n, int *largest, int *second_largest);

* When passed an array a of length n, the function will search a for its largest and second-largest elements, storing them in the variables pointed to by largest and second_largest respectively.

寫 find_two_largest() 函數。陣列做為函數參數找出最大值跟第二大值。

Finish the code below to match the input and output. Pointer is required. 請使用指標。

#include <stdio.h>

void find_two_largest(int a[], int n, int *largest, int *second_largest);

int main(void)
{
	int n, largest, second_largest;
	scanf("%d", &n); //Numbers will be entered
	int a[n];
    //Enter n integers separated by spaces
	for (int i = 0; i < n; i++)
		scanf(" %d", &a[i]);

	find_two_largest(a, n, &largest, &second_largest);

	if (n == 0)
		// Your code here
		//Print No numbers were entered.
	else if (n == 1)
		// Your code here
		//Print Only one number was entered. Largest: 
	else
		// Your code here
		//Print Largest: , Second Largest:

	return 0;
}

void find_two_largest(int a[], int n, int *largest, int *second_largest)
{
	// Your code here (Using pointer to finish)
}

Example input:

Example output:

No numbers were entered.

Example input:

1
67

Example output:

Only one number was entered. Largest: 67

Example input:

5
5 6 7 8 9

Example output:

Largest: 9, Second Largest: 8

[Exercise] Coding (C) - asked Dec 21, 2017 in Chapter 12: Pointers and Arrays
ID: 38251 - Available when: Unlimited - Due to: Unlimited
edited Dec 21, 2017 by semicolon | 18.1k views

58 Answers

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Unknown · Answer 1 · 2018-01-18 17:39:01

Hidden content!
#include <stdio.h>

void find_two_largest(int a[], int n, int *largest, int *second_largest);

int main(void)
{

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}

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{
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}

answered Jan 18, 2018 by Unknown (-140 points)

112233 · Answer 2 · 2018-01-15 09:29:45

Hidden content!
#include <stdio.h>

void find_two_largest(int a[], int n, int *largest, int *second_largest);

int main(void)
{
    
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}

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{
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}

answered Jan 15, 2018 by 112233 (-167 points)

Sabrina Lin · Answer 3 · 2018-01-13 00:45:11

Hidden content!
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{
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}

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}

answered Jan 13, 2018 by Sabrina Lin (-285 points)

PsychoDeity · Answer 4 · 2018-01-10 13:23:07

Hidden content!
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void  *    * ***  a[], int n, int  *  * * int      *     * 

int *  ****
{
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*  *    **   * *   * *  * * ** ***   ****  * **     *    *  *

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* *******     * *   ** *** * * *    //Print Only one number was * *   * ****
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}

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{
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}

answered Jan 10, 2018 by PsychoDeity (-168 points)

prog.c: In function 'find_two_largest':
prog.c:33:17: warning: comparison between pointer and integer
         if(a[i] > largest){
                 ^
prog.c:35:21: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
             largest = a[i];
                     ^
prog.c:36:25: warning: comparison between pointer and integer
         } else if (a[i] > second_largest
                         ^
prog.c:37:21: warning: comparison between pointer and integer
             && a[i] < largest){
                     ^
prog.c:38:28: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
             second_largest = a[i];
                            ^

commented Jan 10, 2018 by The Judge (4.3k points)

410330002 · Answer 5 · 2018-01-04 15:31:20

Hidden content!
#include ***   *** * **
#include  * * *     *     **


void *  **     * * a[], int n, int *largest, int  * * * ** * *

int **    * 
{
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* **   **  ** **  *     ** **  *** *  * ****   
** * *  *   *  int a[n];
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**  *  *  * ** ***   ***   *   *  *****    **  * * * **     numbers were ***** * * *  

* * ** *  **** * else if (n == 1)
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  **  ** *  else
  **   **    * **   ** *** **          *   %d, Second Largest:  * **  *** *** ** *** 

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}

void ** *    ****   a[], int n, int *largest, int    ****** * *
{
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* **    ** ********* *** 
}

answered Jan 4, 2018 by 410330002 (-16 points)
edited Jan 17, 2018 by 410330002

林上智 · Answer 6 · 2018-01-03 16:21:34

Hidden content!
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void *  *     *   *  a[], int n, int   **  * int ** *****  **   

int   * * *
{
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* **     *** *    ***  * ***   n, ** ** ****    ***  * ***  * *   **

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*  * * **  **  *  *   * * * *** // Your code here
***  **** ****  * *  *** * ** //Print Only one number was   *    *
***** * *  **        ** *      *  *     **   one number was *   *  ******   * **  ** 
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 *  ** * ***   **** **** ** * ** **  *** ***  **   * ** * **  **      **   ***  %d, Second *         * ** * ** * ****  * *  *

** *  ** ** **  * return 0;
}

void **  * ** *** a[], int n, int * **** ** int      * *
{
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}

answered Jan 3, 2018 by 林上智 (-32 points)
edited Jan 3, 2018 by 林上智

fung2004067 · Answer 7 · 2017-12-30 21:44:05

Hidden content!
   ** *  ** *** ** ** *


//void  *** ** *  a[], int n, int ** **  int   ** ****   **

int ** ** **  largest, int ** *   * * *
{
******* *  ***  *  return (largest == ** * *    *
}

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{
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** *  * *   * * for(i = 0; i ****  n; i++)
**** *  *   *   ** {
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* *  ***   * ** ****** * **** * * ******* 
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*   * *    *  **  * * **  *****  * *  ****
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*   *   *  * * 
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*  * **** *    *       * }

  **** *  **  *  return 0;
}

answered Dec 30, 2017 by fung2004067 (-214 points)

fung2004067 · Answer 8 · 2017-12-30 21:27:35

Hidden content!
 ** ***   *** *** * * **


//void  ***        a[], int n, int * ***  int  *  **  **


void     *****      a[], int n, int* *  ***  int*   *  *  
{
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** * *   * *   for(i = 0; i  **  * n; i++)
**  *  **  * *  {
**   **  * ***    ** ***   ** ** if (a[i] >  *   **
**  * *    ******     * ***   ****  
*** *       *** ** ** * **   * *  * ****    _largest = a[i];
* * *** * *  ****  ** * * **    * *  * **  *
*    * ***   }
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****    *  ****  * 
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*  ***** **    *    *****   = 0; i < n; i++)
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** *     * ** }
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}


int **  * 
{
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*   ***  * *   *** ** **  **  *** * *    **  *   *   **** ***** * * will be entered
    **  * *** *  int a[n];
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*  ** *  *  int i;
**  *    * ** ** for (i = 0; i *  *** n; i++)
** **  *    *  *** * *****  * 
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****          *  *  **   * * *  n,  * **   **** *   *  ** * **  

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*  **    **     * *** *       *** }
* *  ** **  else if (n == 1)
** *  * ** * *  * **  ** * *   *  {
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  ***  *** *            *     }
*     ** **  else
*  * ** *** * *   * **  * *   * {
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}

answered Dec 30, 2017 by fung2004067 (-214 points)

fung2004067 · Answer 9 · 2017-12-30 20:03:23

prog.c: In function 'find_two_largest':
prog.c:10:11: warning: comparison between pointer and integer
  if (a[i] > largest)
           ^
prog.c:12:16: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
        largest = a[i];
                ^
prog.c: In function 'main':
prog.c:40:14: warning: missing terminating " character
       printf("Only one number was entered. Largest:%d,&largest);
              ^
prog.c:40:14: error: missing terminating " character
       printf("Only one number was entered. Largest:%d,&largest);
              ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.c:41:3: error: expected expression before '}' token
   }
   ^
prog.c:41:3: error: expected ';' before '}' token
prog.c:44:14: warning: missing terminating " character
       printf("Largest: %d, Second Largest: %d,&largest, &second_largest);
              ^
prog.c:44:14: error: missing terminating " character
       printf("Largest: %d, Second Largest: %d,&largest, &second_largest);
              ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.c:45:3: error: expected expression before '}' token
   }
   ^
prog.c:45:3: error: expected ';' before '}' token

~~~~~~ · Answer 10 · 2017-12-28 17:25:24

Hidden content!
#include  * ** *** ** *  
#include ****  * *  ** ***


void *  *  * ** *  *   a[], int n, int *largest, int *  *  ** 
{
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*     ***       * }
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void  *** * **   ***  a[], int n, int *largest, int  *   * *  * * 

int  *  
{


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    * ** ***   **  *    **  ** *      * *
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**** *    for (int i = 0; i < n; i++)
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 *** **** *  **      * ***  n,  *** **    * ****  ***  *


******** *  * return 0;
}

answered Dec 28, 2017 by ~~~~~~ (-304 points)
edited Dec 28, 2017 by ~~~~~~

12-21 Exercise 1: Finish the code to match the input and output. Pointer is required.

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