C Program to Accept n Integers in an Array and Search for a Specific Number

(Motivation) I am trying to construct a program in C that allows me to (1) Enter a certain amount of numbers in an array (2) Enter what exactly those numbers are (3) Enter a search value from that array that makes the C program find what position that search value is in.

Basically, I am trying to do this.

(Attempt) Here is what I tried so far.

#include <stdio.h>
#include <stdlib.h>
//Write a 'C' program to accept n integers in an array and search for a specific number.

int main()
{
    int a[10],n,i,key;
    printf("Enter how many numbers in an array: ");
    scanf("&d",&n);
    printf("Accept n numbers in an array: \n");
    for(i=0;i<n;i++)
    {
        scanf("&d",&a[i]);
    }
    printf("Display Array Elements\n");
    for(i=0;i<n;i++)
    {
        printf("%d",&a[i]);
    }
    printf("Enter search value: ");
    scanf("%d", &key);
    for(i=0;i<n;i++)
    {
        if (a[i]==key)
        printf("Number found at position %d", i+1);
        else
        printf("Element not found!");
    }
    return 0;
}

When I compile this in CodeBlocks (version 20.03), I am able to enter in how many numbers there are in an array, but then after typing in a number, say 4, CodeBlocks prints out

Accept n numbers in an array: 4

Display Array Elements

Enter search value:

Process returned 0 (0x0) execution time : 22.628 s

Press any key to continue.

(Question) How can I improve this code such that it does what (1), (2), and (3) above says? I also welcome alternate methods. I have been trying to replicate what the YouTube video in the link says, but I have not been able to succeed.

I want to mention I have little to no coding experience, and I am learning C for the first time. I also tried searching for similar questions, but given my novice skills in coding, I am not able to take advantage of them. Thank you in advance.

Solution

At least these problems:

Code not compiled with a well enabled compiler

Use a compiler with all warnings enabled. This rapidly points out many of code's troubles.

warning: too many arguments for format [-Wformat-extra-args]
scanf("&d", &n);
      ^~~~

warning: too many arguments for format [-Wformat-extra-args]
scanf("&d", &a[i]);
       ^~~~

warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
printf("%d", &a[i]);

Code attempts to print the address and not the value

Printing with "%d" and an address and not an int leads to undefined behavior (UB). Drop the &.

// printf("%d",&a[i]);
printf("%d", a[i]);

Use % @Eric Postpischil

Since &d is not a specifier, "&d" directs scanf() to scan in the characters '&' and 'd' and then move on. If there or not, code simply moved on and used an uninitialized n for the rest of code. In OP's case, n appears to be 0 and so code ended promptly.

// scanf("&d",&a[i]);
scanf("%d", &a[i]);
//     ^

Test inputs

Check scanf() return value. @Shawn
Test range

// scanf("&d",&n);
if (scanf("%d", &n) != 1 || n < 1 || n > 10) {
  fprintf(stderr, "Invalid numeric input or out of range 1-10\n");
  return EXT_FAILULRE;
}