当我没有动态声明我的数组时，这个 display() 如何显示数组值

Question

From my knowledge after a function is called in c++ its memory is deallocated for another variables. If it doesn't allocates to another variable then variable should have allocated memory dynamically. I'm confused how the function display() displays array values when it isn't allocated memory dynamically.

#include<iostream>
using namespace std;
void init_values(int arr[]){
    for(int i=0;i<100;i++){
        arr[i]=i;
    }
}
void display(int arr[]){
    for(int i=0;i<100;i++){
        cout<<arr[i] << " ";
    }
}
int main(){
    int arr[100];
    init_values(arr);
    display(arr);
}

I expected the function displays garbage or will show an error. But it displayed the values correctly.

Answer 1

void init_values(int arr[]){
    for(int i=0;i<100;i++){ // i is created here
        arr[i]=i;
    } // i goes out of scope here
}

No problems here, there's no attempt to use i later.

void display(int arr[]){
    for(int i=0;i<100;i++){ // i is created here
        cout<<arr[i] << " ";
    } // i goes out of scope here

Again, no problem.

int main(){
    int arr[100]; // arr is created here
    init_values(arr);
    display(arr);
} // arr goes out of scope here

When we call init_values and display , arr is still in scope since its lifetime ends at the end of the scope it was created in. So it's still in scope during those function calls.

Now, this would be a problem:

int *init_values(void){
    int arr[100];
    for(int i=0;i<100;i++){
        arr[i]=i;
    }
    return arr;
}

Why? Think about it:

int *init_values(void){
    int arr[100]; // arr is created here
    for(int i=0;i<100;i++){
        arr[i]=i;
    }
    return arr;
} // arr goes out of scope here

So the caller of init_values would receive a pointer to arr 's contents even though arr does not exist after we exit this function. This function's return value could not be safely used. Your code doesn't do this.