Question

Consider the following Python program: def fun(x, y): return x + y # [2] # [1] a = fun(2, 3) b = fun("2", 3) print a, b What does it evaluate to? Replace the last statement print a, b with print a + b and explain the traceback. What's wrong? Now eliminate the line marked [1] and change line [2] to read return x + y. Run the program and explain the traceback. Consider the following definition: def fun(n, m): return m - n Evaluate the following expressions (perhaps using stack diagrams: a) fun(fun(1, 2), 3) b) fun(fun(1, 2), fun(3, fun(fun(4, fun(5, 6)), 7))) c) fun(fun(1, 2), fun(3, fun(fun(4, fun(5, 6)), fun(7, 8)))) What happens if in the definition of fun above we replace return by print? Considering the following definitions: def alpha(x, y): return x + beta(y, x) def beta(x, y): return y - x # [1] What does alpha(2, 3) evaluate to? How does the answer change if the line marked [1] is changed to return x - y?

Answer 1

Here is the explanation of how all the above functions works along with the screenshots of code and output.

1. def fun(X,Y):

return X*Y

a = fun(2,3)

b = fun("2",3)

print(a , b)

Explanation :

- 1. def fon (x, y): return xxy a : fon (2,3) be fun ("2",3) print (a,b) Solution : 16 "222" How ? 1. When the program starts it will forot go to a = fun (2,3) - 0 Then 'fun with arguments 2, 3 will be called entust 12,3! - fun (2, 3] x def fun (x, y). 1; As both X & Y are of return 2x3 int type product take int place Now a = fun (2, 3) → Returned value This 6 will be stored in a Now a = 6

2. After a = ton (2,3), the interpreter goes to b = (23) string int "2" 3 Now fon ("2", 3) will call def for (x, y) ton ("2",3) n det fun (x,y) which will return "2"* 3 + return "2-3 times 922. a t The int(3) will be considered as number of times i the string should 7 be itepeated in til .. .. fon ("2", 3) + det fun (x,y) booter return "2"*3 Now bu fun ("2", 3) will assign result of fun ("2", 3) to b. now b="222".

3. The cursor moves to print (a,b). As the two variables are seperated by () comma, they will be printed as it is. Output : 6222|| i hely y)

SCREENSHOT(CODE AND OUTPUT)

When you run the above code the output looks like below.

1 def fun (X,Y): return X*Y s L a = fun (2, 3) 6 b = fun("2",3) P CÓ print (a, b) 8 9 10 Shell >>> %Run Python1.py 6 222 >>>|

1(b) . If we change print(a , b) to print(a + b) , then it will show an error because

The output of fun(2,3) which is in a is of int type ( a = 6) The output of fun("2",3) which is in b is of string type.(b = "222") When we try to print a + b as string cannot be concatenated with int directly it will show an error

SCREENSHOT

def fun (X,Y): return X*Y a = fun(2, 3) HNM to 00 b = fun("2",3) print (a + b) 10 Shell >>> %Run Python1.py Traceback (most recent call last): File "C:\Users\Neelima Rambabu\Desktop\Pythonl.py", line 8, in <module> print (a + b) TypeError: unsupported operand type (s) for +: 'int' and 'str'

1(c) If we remove a = fun(2,3) and change return X*Y to return X + Y it will also generate an error . To understand this see the below trace back.

1.c) def fun (x, y) return x + y d ba: fon("2", 3) print (a + b) solution ; "Error. 1. The program starts at Now, fon ("2", 3) will 16for ("2", 3)-6 call det for (x,y) et turl y ) stringentint "2" 333 Bio t. fon ("2", 3) - det forl x,y) Treturn x + y ) "2" 3 This is the part where erro Occors and program cannot go further.

because, X = "2" is of string type, Y = % is of integer type So, x + y is not possible as string came be concatenated directly to integer.

SCREENSHOT

def fun(X,Y): return X + Y Homto b = fun("2",3) 8 print (a + b) 10 Shell >>> %Run Python1.py Traceback (most recent call last): File "C:\Users\Neelima Rambabu\Desktop\Pythonl.py", line 6, in <module> b = fun ("2", 3) File "C:\Users\Neelima Rambabu\Desktop\Pythoni.py", line 2, in fun return X + Y TypeError: can only concatenate str (not "int") to str >>>|

2(a) The explanation in the form of stack for 2 (a ) is as follows

☺ det fun (nm): return mo a) fun ( fun (1, 2), 3) When the interpreter goes 'to ton ( fun (1, 2), 3) it will evaluate inner function feret i.e., 'fun (1, 2) and takes its return value and pass it as an argument to outer fonction. :. (1,2)= det fon inim) - "return man fun (fun (1,2),3) tunlin13) det fun (1,3) reton 3-1..) 2.

:: So, finally the output of fon ( fun (1,2), 3) will be 21. This flow of execution takes place for every function ie; (b, c) given. When we reprennt it po Stack ! fur (fun (1,21,3) as top value became value than function stack will pop stems S--cccn. fun (1,2)= 1 @ . ..>fun (fun (1,2), 3) * Output | 2 fun (1,3) B + À →

For 2(b) & 2(c)

Explanation

26 fun (fun (1,2), funt 3, fun (4) fur (5,6)), - ))) The execution flow of this will also be as same as above with some differences in Stack. O p ben mes E ant fon (5,6)= 1 top becomes Integer, stack • traces back fun(fun (1,2), fun (3 fulfon (4, for i (5,61), 7)). i o R oute' o . fun (4, tun(5,6) fon (3 ton (4, fun(5,6), +) fon for (i, tun( 3 Pornca turlsang a fun (1,2)=1 fun (fun (1/2), fun(3, funk (fun (4, fon (516)),7) from from other back Stack in memory 0-6 functions are stored in stack the top becomes int and as no function to store it well trace evaluating all functions as follows.

= 1 m n + 6-5 m D * 1 - 4 ( 8 o foo( 4 m ) = -3 a fun (-3,7) →7-(-3) = 10 0 ton ( 3 , 10) = 10-3 = 7 = 6 + Final result. w fun (1,7) 2. © The- ah fun (tun (1,2), fun (3, fon (fun (4, fun (5,6)), fun (7,8)))). The above stack opplies same for this with one change for fun (4,8) as follows As the explanation is same for 2(c) except at fun (7.8) ; would like to give it as below",

om Mo The 7 will have fun ( 7,8) = 8-7. → tunaton ( 4, tun (5, 6)), ton (7,8 )) funcs,6) - 6-5 = 1 9 → tun (4,1) = 1-4 -3 io → tun (-3, 1) = 1-(-3) = 4 ĥ → fun (3,4) = 4-3 = 1 12 → func fun (1,2), 1) fupce fun (12) = 1 . 13) fun (e,t) ali =0 → Final rewit

SCREENSHOT

def fun(n,m): return m - n Hem tin print("fun(fun(1,2),3) = {0}".format(fun(fun(1,2),3))) 6 print("fun(fun(1,2), fun(3, fun(fun (4, fun(5,6), 7)) = {0}".format(fun(fun(1,2), fun (3, fun(fun (4, fun(5,6)), 7 ) >>>> 8 print("fun(fun(1,2), fun(3, fun(fun (4, fun(5,6)), fun(7,8)))) = {0}".Format(fun(fun (1,2), fun(3, fun(fun (4, fun(5,6)), fun(7,8)) Shell >>> %Run Python2.py fun (fun (1,2), 3) = 2 fun (fun (1,2), fun (3, fun (fun (4, fun (5,6)), 7))) = 6 fun (fun (1,2), fun (3, fun (fun (4, fun (5,6)), fun (7,8)))) = 0 >>>|

For third one the output and explanations are as below.

113) fun Cicl) 1-1=0 Final rewlt. ☺ def alpha (x,y): I retun x + beta (x,x) def beta (x,y) : returo y - x. of execution For alpha (2, 3) the flow will be as follows.

2,3) step - step - step - det alpha (x,y) *= 2, Y = 3 x + beta (yp, x) 2 + beta (3,216 beta (3,2) Ly return 2-3 =(-1) step - Step-6 2 +62) = 1 Step - return 1. The output for alpha (2,3) = 1 Foi I changed return Y- to x-y then, The steps from ☺ to 9 are exactly as above but at step - (+1) will be returned as 3 - 2 = 1 x - y step - 2 + 1 - 3 Step - ©: return 3 – Fenal Output

SCREENSHOT

def alpha(X, Y): return X + beta(Y,X) Humin def beta(X, Y): return Y - X 7 print(alpha(2,3) Shell >>> %Run Python3.py 1 >>>|

After changing return Y - X to X - Y

def alpha(X, Y): return X + beta(Y,X) o u AwNN def beta(X, Y): return X - Y 7 print(alpha(2,3) Shell >>> %Run Python3.py 3 >>>|