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Python Python Basics Functions and Looping Exception Flow

Why does it print A, D, F?

I don't understand why it prints A, D, F?

print("A")
try:
    result = "test" + 5
    print("B")
except ValueError:
    print("C")
except TypeError:
    print("D")
else:
    print("E")
print("F")

2 Answers

Steven Parker
Steven Parker
231,275 Points

Since you know the answer, you should have seen this explanation when you selected it in the quiz: "Adding a string and an int raises a TypeError. Since an exception occurred the else block doesn't run."

If we turn that explanation into line-by-line comments:

print("A")               # this prints the "A"
try:
    result = "test" + 5  # adding a string and an int raises a TypeError
    print("B")           # so this line is skipped over
except ValueError:       # this line is also skipped over
    print("C")           # this line is also skipped over
except TypeError:        # the code resumes running here
    print("D")           # this prints the "D"
else:                    # since an exception occurred, the else block doesn't run
    print("E")           # so this line is skipped over
print("F")               # finally, this prints the "F"
Rick Gleitz
Rick Gleitz
47,878 Points

Hi Svenvanzijl,

A and F are going to be printed in any case because they are outside the try/except/else portion of the code. The other choices are within the try/except/else part, so the only thing to figure out is which one. If the value of the variable result was valid, it would print B also (it is not valid). ValueError as I understand it is mostly for catching things that don't normally make mathematical sense, like taking the square root of a negative number, which is not what we have here, so it doesn't print C. It does print D because result tries to add two things that are mismatched types (an integer and a string). Lastly, it doesn't print E because it did catch an error. So it prints A, D, F.

Hope this helps.