1. Suppose we want a program to request an integer (n) repeatedly until it is in
ID: 3889911 • Letter: 1
Question
1. Suppose we want a program to request an integer (n) repeatedly until it is in the range 0 to 100 inclusive. Which of the following is the best input validation loop?
HINT: see DeMorgan’s law document.
A. while n < 0 or n > 100:
generate an error message
B. while n >= 0 and n <= 100
generate an error message
C. while n >= 0 or n <= 100
generate an error message
D. while n <= 0 or n >= 100
generate an error message
2. What sequence of values will be printed when the following instructions are executed?
X = 5
if (X < 7):
print(6)
Y = 6
else:
print(4)
Y = 4
if (Y < 5):
print(3)
else:
print(2)
3. What would be printed if the following instructions were executed?
X = 4
while (X > 0):
print(X)
X = X – 1
4. What would be printed if the following function were executed with the value of N being 9
def xxx(N):
if (N < 7):
print(N)
else:
N = N + 3
print(N)
Explanation / Answer
1) Correct condition should be
while ~ (n>=0 && n<=100)
Applying demorgan law ::
while n<0 or n>100
Answer is A. while n < 0 or n > 100:
2)
X = 5
if (X < 7):=> TRUE because X = 5
print(6) =====> Prints 6 because X<7
Y = 6=====>Y becomes 6
else:
print(4)
Y = 4
if (Y < 5): ==> False because Y = 6
print(3)
else:
print(2) ==> Finally Prints 2
Output : 6 2
3)
X = 4 ==> Initially X is 4
while (X > 0): ===> LOOP WILL RUN UNTIL X> 0
print(X) ==> Prints 4 , 3, 2, 1
X = X – 1
When X becomes 0 , loop terminates
4)
N is 9
def xxx(N):
if (N < 7): ==>False because N = 9
print(N)
else:
N = N + 3 =====> N BECOMES N +3 = 9 +3 = 12
print(N)==> Prints 12
OUTPUT : 12
Thanks, let me know if there is any doubt
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