Why would my array would be filled out to zero, when I initialised it to -1

Question

#include<iostream>
using namespace std;

long long int memo[20] = {-1};       //create memo table  and initialise to -1

long long int fibo(long long int n)
{
    if(memo[n]>-1)           //changing this to if(memo[n]>0) works fine
        return memo[n];      //but as such this gives 0 from all my inputs
    if(n<=2)
        return 1;
    memo[n] =  fibo(n-1) + fibo(n-2);    //recurse
    return memo[n];
}

int main()
{
    long long int n;
    cin>>n;
    cout<<fibo(n)<<endl;       //calls the fibo function 
    for(int i=0;i<20;i++)       //this prints my memo table used...
        cout<<memo[i]<<" ";
}

I am calculating the nth Fibonacci number using top-down dp but my memo table is zeroed out. Even at locations which I am not touching, why?

Answer 1

When you have

long long int memo[20] = {-1}; 

You are not telling the compiler to initialize the array with all -1. What you are telling it is here is a initializer list and initialize the elements in the array with its contents. Since the list is smaller than the array every missing initializer causes the compiler to zero initialize the corresponding elements.

Answer 2

Because that's how array initialization works in C++. You set the first element of the array memo to -1, and the compiler will value-initialize (before the C++11 standard) or default-initialize (since C++11 and onward) all of of the other elements.

Please read more about aggregate initialization here.