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4 votes
Given the following code snippet:

SecCond = "SS
SecStatus = false
try (if (SecStatus)SecCond = 2SS else
SecCond = normal operations (nothing to see here)} catch (e) {
SecCond = normal operations (e) (Exeption logged)}

Which of the following failure modes would the code exhibit?

A. Open
B. Secure
C. Halt
D. Exception

User Thotruong
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7.5k points

1 Answer

3 votes

Final answer:

The code exhibits an 'Exception' failure mode if an error occurs triggering the catch block to execute.

Step-by-step explanation:

The provided code snippet contains a try-catch block with conditional logic based on a variable SecStatus. Based on the code, if SecStatus is true, SecCond is set to '2SS'; otherwise, it defaults to 'normal operations (nothing to see here)'. However, if an exception occurs during the execution of this code, the catch block will execute, setting SecCond to 'normal operations (e) (Exeption logged)'. Given the options provided, the code would exhibit an Exception failure mode if there is an error during the execution of the try block that triggers the catch block.

The code in the snippet is written in JavaScript and contains a try-catch block. The try block attempts to execute some code, and if an exception occurs, the catch block is executed. In this specific code snippet, the failure mode that would be exhibited is an Exception. This is indicated by the catch block where the value of SeCond is set to "normal operations (e) (Exception logged)". This suggests that an exception occurred and it has been logged.

The other failure modes mentioned in the options are not present in the code snippet. There is no indication of an "Open" failure mode, a "Secure" failure mode, or a "Halt" failure mode.

User James Roland
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7.0k points