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i = 0
print(f"initial-value: {i}")
while i != 0:
i = i - 1
print(f"while-value: {i}")
else:
i = i + 1
print(f"else-value: {i}")
print(f"final-value: {i}")
initial-value: 0
else-value: 1
final-value: 1
The answer should be A because i is 0 in the first line, which means While won't be execute, then direct goes in to else section. i = i +1 which means i = 0+1, the result is 1.
The value of the i variable when the while loop finishes its execution will be 1. This is because the else clause of the while loop will be executed only if the condition of the loop (i != 0) is not met, which means that the loop will not be executed at all. Since the loop is not executed, the value of i will not be changed, and it will remain 0. However, the else clause will be executed, which contains the statement i = i + 1. This will set the value of i to 1, which is the final value of i when the loop finishes its execution.
is it possible that the answer is D, the variable becomes unavailable? Technically the else statement
is: 0 = 0 + 1, we are saying nothing is equals to something? I know when we run the code and add a print statement the returned value is 1, but the code snipped does not include a print statement, so when the code is run without a print statement, we have to assume the answer is D noting is returned?
A is correct.
First i is given the value of 0.
The next line says "if i is not equal to 0, then decrease i's value by 1". The value of i is currently 0, so this condition does not apply and the programme proceeds to the "else" part, which is "add 1 to i". The value of i is 0, so adding 1 gets you 1.
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