Why memoization doesn't require a pointer?

Question

For example in this code, which works fine, memoization done with an array not a pointer to an array. (int mem[] instead of int* mem)

int fibonacci(int n,int mem[]){
    if(n < 3)
    return 1;   

    if(mem[n])
    return mem[n];

    mem[n] = fibonacci(n - 1, mem) + fibonacci(n - 2, mem);
    return mem[n];
    } 

int climbStairs(int n){
    int mem[50] = {0};
    return fibonacci(n + 1, mem);
}

I don't see how, when one of the functions returns, how can the caller function of the returned function have the data to previous mem[n]. Isn't the functions array supposed to be deleted when the function returned? How can after that function returned, another function that a caller function calls can have that value?

Note: This is a solution to Climbing Stairs in leetcode. Written by me.

See https://stackoverflow.com/questions/1461432/what-is-array-to-pointer-decay — Vlad Feinstein, Oct 05 '20 at 22:41
it is passed as a pointer in fibonacci. memory is allocated on the 'stack' in climbStairs — Abel, Oct 05 '20 at 22:42
https://stackoverflow.com/a/6567846/3365922 because `int mem[]` and `int* mem` are same — fas, Oct 05 '20 at 22:42

score 0 · Answer 1 · answered Oct 05 '20 at 22:42

0

When using [] in a function argument, it is a pointer.

int foo(int x[])

and

int foo(int *x)

are 100% equivalent.

However, this would be something else:

int foo(int x[10])

answered Oct 05 '20 at 22:42

klutt

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Why memoization doesn't require a pointer?

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