Function Print1() puts "First" and Function Print2() puts "Second". Print1() is called earlier. What is put to output by calling Print2()?

Rationale

Print2() outputs "Second".

When Print2() is called, it executes its defined action, which is to output "Second". The earlier call to Print1() does not affect the output of Print2() since each function operates independently.

A (FirstSecond) YOUR ANSWER

This choice incorrectly combines both outputs from Print1() and Print2(). However, since Print1() is not called during the execution of Print2(), there is no concatenation of outputs, making this choice incorrect.

B (Second) CORRECT

This is the correct answer as Print2() directly outputs "Second" when invoked. The function's definition specifies what is to be printed, and there is no interference from any previous function calls.

C (First) YOUR ANSWER

This choice suggests that Print2() would output "First", which is incorrect. Print1() outputs "First", but it does not influence the output of Print2(), which is specifically programmed to output "Second".

D (Second First) YOUR ANSWER

This option implies that "Second" would be printed first, followed by "First". However, since Print2() does not call or interact with Print1(), it solely outputs "Second", making this answer inaccurate.

Conclusion

The output of the function Print2() remains independent of previous function calls, such as Print1(). Since Print2() is designed to output "Second", this is the only text displayed when it is called, highlighting the independence of function execution in programming.

Function Print1() puts "First" and Function Print2() puts "Second". Print1() is called earlier. What is put to output by calling Print2()?

Print2() outputs "Second".

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