Question

Suppose you were debugging the push() function of your program. Which of the following variables would be accessible to the debugger before the function is called? static struct node *stack; static struct node *new_node() { int size = sizeof(struct node); return malloc(size); void push(void *value) { struct node *n = new_node(); n->value = value; n->next = stack; stack = n;

Answer 1

1) 3rd option - stack, node, value and size.

Because, push() is the function which is used to insert or append a new element into the stack. New element is inserted by creating a new node to store the value and after inserting the value or data into the node, stack pointer will move to next address.

Here, before using push(), a new node is created to store the value in stack. Size is allocated dynamically bu using malloc function. So, the variables accessed before using push() are stack,node,value and size.

So, 3rd option is correct.

2) 1st option - 3.

When display (foo(counter)) is called, the counter value is obtained from the definition counter(new-count). Here, counter(new-count) holds the value obtained from new-count(lambda()). Initially, cnt is 0 and incremented if the condition is true. So, cnt value increase for 3 times. Hence, cnt hold value of 3. This value is finally returned to foo(counter). Hence display () function gives output as 3.

3) 1st option - (-1).

When print() is called, foo() is called. Now, fun is 0. Then, this foo() function holds the value fun+bar(fun-1), i.e foo() holds 0(+bar-1). Where, bar(-1) holds the value fun+1. Where, fun is -2. So, bar(-1) holds -1. Finally print (foo()) display output as -1.

Comment if u have any doubt.

Thank you.