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abruzzese [7]
3 years ago
15

Consider the following pseudocode:

Computers and Technology
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:

Part a:  The program will print 3.

Part b: The program will print 4

Part c: The program will print 1.

Explanation:

Part a: As the scoping is static, the x integer has a value of 1 and the y integer has a value of 2 so the addition is 2+1=3.

Part b: As the scoping is dynamic with deep binding, the x integer has a value of 2 and the y integer has a value of 2 so the addition is 2+2=4.

Part c: As the scoping is dynamic with shallow binding, the x integer has a value of 1 so it will print 1.

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