Initialising array elements in curly braces???

Question

Specifically in the code below, what's the difference between the first way and the second way?

#include <iostream>
using namespace std;

int main() {
int arr[] = {0,-1,2};
arr[0]={0-arr[1]}; //First way
cout << arr[0] << '\n';
arr[0]=0-arr[1]; //Second way
cout << arr[0] << '\n';
return 0;
}

Answer 1

from the following post i see it has to do with implicit conversion of airthmetic values (i.e. double to int which will generate warning)

https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.educative.io/answers/declaring-a-variable-with-braces-in-cpp&ved=2ahUKEwjSyP6g_Mr4AhXxjOYKHZR4Ay8QFnoECBcQAQ&usg=AOvVaw2kNjVySXVNbtsFuMZOoIZU

Answer 2

assignment using curly braces prohibits <implicit narrowing conversions> among built-in types.
for example if the value of an expression in a curly braces isn’t guaranteed to be expressible by the type of the object being initialized or assigned, the code won’t compile:

https://godbolt.org/z/xfWMbKoEe

however assignment using parentheses doesn’t check for narrowing conversions:

https://godbolt.org/z/GvsKz9KqG

have a look at this article by herb Sutter:
https://herbsutter.com/2013/05/09/gotw-1-solution/