why this code ask two input and then run please help

Question

#include<stdio.h>

struct Array {

int A[10];
int size ;
int length;
};

void Display(struct Array arr)
{
for(int i=0; i<arr.length; i++)
{
printf("%d \n\n",arr.A[i]);
}
}

int Search(struct Array Panda,int a) {

for(int i=0; i<Panda.length; i++) {

if(a==Panda.A[i])

return i+1;
}
return -1;

}

int main()
{
int a;
struct Array panda= {{2,3,4,5,6},10,5};

Display(panda);

printf("Enter number to search : ");
scanf("%d \n",&a);

printf("Found at %d position ",Search(panda,a));

//If we get -1 as return then
//it means number not found

return 0;
}

Answer 1

No it's asking only one input. try again

Answer 2

No brother after we entered which number we had to search if we chose example 4 then we need to enter another number to run code
Answer is correct but why we need to enter one another number

Answer 3

#include<stdio.h>

struct Array {

int A[20];
int size ;
int length;
};

void Display(struct Array arr)
{
for(int i=0; i<arr.length; i++)
{
printf("%d \n\n",arr.A[i]); //this displays 5 elements of A[] array if structure
}
}

int Search(struct Array Panda,int a) {

for(int i=0; i<Panda.length; i++) { //Panda.length=5 you assigned

if(a==Panda.A[i])

return i+1; //this is position of element index+1
}
return -1;

}

int main()
{
int a;
struct Array panda= {{2,3,4,5,6},10,5};//here you have length variable assigned 5

Display(panda); //displays 5 elements of A[] array

printf("Enter number to search : ");
scanf("%d \n",&a); //this is only only you are asking ..

printf("Found at %d position ",Search(panda,a)); //this will Search and return a value

//If we get -1 as return then
//it means number not found
return 0;
}
//i enter 4, result is found at 3
//i enter 7 result is -1
//even if you have {1,2,3,4,5,6,7},for input 7 it returns -1