在堆栈上分配的作用域结构

Question

I am reading and learning on C. I have read many similar questions but most of them seem to be a counter example of what I am experiencing or I still don't understand the concept of object allocation on the heap and on the stack.

Suppose I have a struct like in the following example:

typedef struct {
    int x;
    char* word;
    struct list_element* next;
}list_element;

I want to write a function that initialises a struct of the type list_element . What I would do from what I learned from text books is create the struct on the heap using malloc so everything is still visible outside the initialiser function.

list_element* init_list_element(int x, char* word) {

    list_element* le = (list_element*)malloc(sizeof(list_element));
    le->word = word;
    le->x = x;

    return le;
}

For truly understanding the subject I tried to write an alternative initialiser function that allocates the struct on the stack. Which should result in an error later when trying to access the attributes because the variable should have gone out of scope.

list_element init_list_element(int x, char* word) {
    list_element le;
    le.word = word;
    le.x = x;

    return le;
}

However when creating a struct with the second implementation and trying to access for example the attribute x the code doesn't break. Why is this? Shouldn't the variable le be gone out of scope and therefor inaccessible when trying to printf its attributes?

list_element test = init_list_element(123,"test");
printf("%s, %i", test.word, test.x);

Answer 1

list_element test = init_list_element(123,"test");

You are creating a copy of the struct created inside the second function, therefore you can still access it, but it's just a copy.

In C, everything you assign to variable is a copy.

Example:

list_element le1 = {1, NULL, NULL};
list_element le2 = le1;

le2 will contain a copy of le1 and not the original le1. In other words, the following will not change the value of le1:

le2.x = 3;
printf("%d",le1.x); // 1
printf("%d",le2.x); // 3

Answer 2

The variable le has gone out of scope, but you are not accessing le . The init_list_element() function returns a copy of le , and the contents of that copy are assigned to test .

The members of test then contain the same data that le originally did.