Exercise 8-2 Find the specified element in the array (15 points)

This question requires the implementation of a simple function to find a specified element in an array.

Function interface definition:

int search( int list[], int n, int x );

Among them list[]is the array passed in by the user; n(≥0) is list[]the number of elements in the middle; xis the element to be searched. If found

The function searchreturns the minimum index of the corresponding element (the index starts from 0), otherwise it returns −1.

Sample referee test procedure:

#include <stdio.h>
#define MAXN 10

int search( int list[], int n, int x );

int main()
{
    
    
    int i, index, n, x;
    int a[MAXN];

    scanf("%d", &n);
    for( i = 0; i < n; i++ )
        scanf("%d", &a[i]);
    scanf("%d", &x);
    index = search( a, n, x );
    if( index != -1 )
        printf("index = %d\n", index);
    else
        printf("Not found\n");

    return 0;
}

/* 你的代码将被嵌在这里 */

Input example 1:

5
1 2 2 5 4
2

Output sample 1:

index = 1

Input example 2:

5
1 2 2 5 4
0

Output sample 2:

Not found

answer:

int search( int list[], int n, int x )
{
    
    
	int *p = list; //定义指针p指向list的首地址
	int i; //循环变量
	int index = -1; //相应元素的最小下标
	
    /*循环遍历数组，寻找有无相应元素*/
	for (i = 0; i < n; i++, p++)
	{
    
    
		if ( *p == x) //如果找到下标给index，退出循环
		{
    
    
			index = i;
			break;
		}
	}

    return index; //返回index的值给search

}