int foo(){
int fun = 0 ;
int bar(int fun){
fun = fun - 1 ;
return fun-1 ;
}
return fun + bar(fun + 1) ;
}
print(foo())
output of code = -1 (option-a)
Question 6 (1 point) Consider the following piece of code. What will the output be of...
Consider the following piece of code int adjust = 0; int foo(int factor, function qux) { int offset = bar(factor) + adjust adjust = (qux (offset) + adjust) / 2; return offset; } Which of the names (symbols) are bound at run-time? factor, qux, and offset foo, factor, and qux All the symbols. adjust, factor and offset
Question 6 (1 point) What is the output of the following code snippet? Please ensure your answe. spaced according to the program code output. public static void main(String[] args) int channel = assignChannel (2); System.out.println("Channel: " + channel); ) public static int assignChannel(int channel) return channel + 3; > A/
Give the output for the following code assume int globalvar 10: above the function prototypes void foo(int&); void bar(int); int main() { int globalVar = 5; foo(globalVar); cout << globalVar << endl; bar(globalVar); cout << globalVar << endl; return 0; } void foo(int& x) { x = globalVar + x; cout << globalVar << endl; return; void bar(int x) { globalVar + x; cout << globalVar << endl; X = return;
Problem 4 (25 points). What will the following code print? Why? Identify the line where the vulnerability is. Explain the vulnerability. Show how the vulnerability can be eliminated Algorithm 2 Example vulnerability l.void called (int foo) 2. if (foo 1) printf("foo") 3. else printf("bar" S.int main O 6. called(2): 7. return 0:)
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; return malloc(size); void push (void *value) { struct node *n = new_node(); n->value = value; n->next = stack; stack =...
Consider the following small program. What will the output be? def foo(x): print(x) def bar(): return 3 def baz(y, z): return y * z a = bar() + 2 b = baz(a, 2) foo(a + b) 10 3 2 25 none of these 15
D Question 3 For the following question, assume that int[] = {6, 2, 4, 5, 2, 1, 6, 2, 5) and consider the recursive method foo" public int foo(int[], int b, int 1) { if ( < length) { if (*[3] !- b) return foo(a, b, +1); else return foo(a, b, 3+1) + 1; }else return ; What is the result of calling foo(a, 3,0);? 0 1 02 3 00
Question 23 (1 point) What is the output of the following code snippet? int[] myarray = {62, 33, 5, 27, 41, 8, 39, 9); System.out.println(myarray[2]); System.out.println(myarray[6]); Blank # 1 Blank # 2 A/
QUESTION 1 What is the output of the following code snippet? int main() { bool attendance = false; string str = "Unknown"; attendance = !(attendance); if (!attendance) { str = "False"; } if (attendance) { attendance = false; } if (attendance) { str = "True"; } else { str = "Maybe"; } cout << str << endl; return 0; } False True Unknown Maybe QUESTION 2 What is the output of the following code snippet? #include <iostream> #include <string> using...
2. Consider the following programs (using C's syntax): #include <stdio.h> int a- 1, b-2; int foo (int x) 1 return (x+a); void ba r () { printf("%d, %d\n",a,b); void foobar) } printf("%d, %d\n", a, b) ; int a -4; bar bfoo (b); bar int main)( int b 8; foobar printf ("%d, %d\n", a, b) ; return 0; (a) What does the program print under static scoping? (b) What does the program print under dynamic scoping?