The program output = 10 30
Shallow binding binds the environment at the time the procedure is
actually called.
It is dynamic scoping with shallow binding
Question 7 (1 point) Suppose the program below outputs 10 30. The language uses int vari...
Consider the following code: int a:=10 //global int b:=12 //global proc F a:= a-b proc P (M:proc) int a:=2 M() proc K int b:=3 P(F) K() //main program print(a) //built in function a- what does this code print if it uses dynamic scoping and deep binding? b- what does this code print if it uses dynamic scoping and shallow binding?
Consider the following pseudocode which uses dynamic scoping. What does the program print if the language uses shallow binding? What does it print with deep binding? x: integer//global procedure print_x write_integer (x) procedure first x:= x * 3 procedure second (F: procedure) x: integer x: = 5 F () print_x () x: = 7 second (first) print_x ()
Consider the following program: # include <iostream> int x = 3, y = 5; void foo(void) { x = x + 2; y = y + 4; } void bar(void) { int x = 10; y = y + 3; foo( ); cout << x << endl; cout << y << endl; } void baz(void) { int y = 7; bar( ); cout << y << endl; } void main( ) { baz( ); } What output does this program...
Can figure out how get my program to keep track of a total score, here is my code so far in python 3.6.6 import math import random import turtle def target(): """Turtle drawing the target""" t = turtle.Turtle() wn = turtle.Screen() wn.bgcolor("black") t.hideturtle() t.speed(0) #most outside circle worth 10 points t.setposition(0,-275) t.color("grey") t.begin_fill() t.circle(275) t.end_fill() #2nd most outter circle worth 20 points t.penup() t.setposition(0,-200) t.pendown() t.color("red") t.begin_fill() t.circle(200) t.end_fill() #3rd most outter circle worth 30 points...