Why my array will be filled to zero when I initialized 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 calculate the nth Fibonacci number using down-down dp, but my note table is reset. Why in places that I don’t touch, why?

Answer 1

Because how array initialization works in C ++. You set the first element of the array memoto -1, and the compiler will be value-initialize (before the C ++ 11 standard) or default-initialize (starting from C ++ 11 onwards) of all other elements.

Read more about aggregate initialization here .

Answer 2

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

-1. , , . , .

Answer 3

long long int memo[20] = {-1}; -1 . 0.

, , , 0, -1 ,

long long memo[20] = {}; /*new C++ standards allow this, note that C doesn't*/

.

Answer 4

-1, {-1}.

++ Stroustrup Array Initializer:

---

, 0 . :

int v5[8] = { 1, 2, 3, 4 };

int v5[] = { 1, 2, 3, 4 , 0, 0, 0, 0 };

---

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

Answer 5

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

, memo .

"memo", , ++. - , memo [3] - - "memo + 3", .

, -1.

long long int memo[20] = {[0 ... 20] = -1};