线性搜索代码表明我的商品不存在。请帮助我进行更正

Question

This is the function for linear search where i am only taking one variable x that is the to search for item variable

int lsrch(int x)
    {int i;
    int arr[6] = {2,4,5,76,2,1};
    for(i=0;i<5;i++)
    {
        if(x==arr[i])
        {
            return i;
        }
        else
            return -1;
    }
    }

    int main()
    {
        int a,b;
        a=lsrch(76);

76 is present so it should show its index location but it shows -1 for both meaning both are not present true for 2nd test case

        b=lsrch(99);
        printf("%d",a);
        printf("%d",b);
    }

Answer 1

Logical Error in your code - the foloowing part of your code was incorrect -

  if(x==arr[i])
   {
       return i
    }
    else 
         return -1

At the first pass itself if condition evaluates false and -1 is returned.

Correct Code -

int lsrch(int x)
   {
    int i;
    int arr[6] = {2,4,5,76,2,1};
    for(i=0;i<=5;i++)
   {
      if(x==arr[i])
      {
        return i;
      }

   }
       return -1;
   }

  int main()
  {
      int a,b;
      a=lsrch(76);
  return 0;
  }

Answer 2

The problem is that you're breaking out of the loop too early.

int lsrch(int x)
{   
    int i;
    int arr[6] = {2,4,5,76,2,1};
    for(i=0;i<5;i++)
    {
        if(x==arr[i])
        {
            return i;
        }
        else
            return -1;      // Incorrect
    }
}

As written, as soon as your code finds a number that doesn't match x , it will return -1. It will never proceed to check the rest of the numbers in arr .

If you compile with gcc -Wall -Werror , the compiler will point out that you've made a mistake:

linsearch.c: In function ‘lsrch’:
linsearch.c:17:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^

This means that you're not returning anything in the case that the loop finishes - causing undefined behavior.

---

The solution is to postpone the return -1 until after the loop has exhausted all of the values in arr .

Also, your loop terminates when i == 5 but you've not checked the last number in arr . Let's use a macro to avoid having to hard-code this value.

#define ARRAY_LEN(x)    (sizeof(x) / sizeof(x[0]))

int lsrch(int x)
{   
    int i;
    int arr[] = {2,4,5,76,2,1};
    for(i=0; i<ARRAY_LEN(arr); i++)
    {
        if(x==arr[i])
        {
            return i;
        }
    }

    return -1;    // Nothing left to check
}

Answer 3

int lsrh(int x){
    int i,a[6]={2,4,5,76,2,1};
    for(i=0;i<6;i++){
        if(a[i]==x)
             return i;
        }
    return -1;
}

use this it will work