Question

(4 pts) What does the function fun print if the language uses static scoping? What does it print with dynamic scoping?

x: integer <-- global procedure

Assignx(n:integer)

x:= n

proc printx

write_integer(x)

proc foo

Assignx(1)

printx

procedure boo

x: integer

Assignx(2)

printx

Assignx(0)

foo()

printx boo()

printx

Answer 1

Answer;

For static scoping:

It will print:

1

1

2

2

Explanation:

First Assignx sets global x as 0.

Outputs:

Line 1: foo() is called which assign 1 to the global x and then prints it i.e. 1

Line2: printx is called which prints the global x which was set to 2 in the previous step

Line3: boo() is called which assigns 2 to the global x by calling Assignx and then prints it i.e. 2

Line 4: printx is called which prints the global x which was set to 2 in the previous step

For dynamic scoping: (function call stack is checked for latest scope)

It will print:

1

1

2

1

First Assignx(0) sets global x since no function is called yet

Outputs:

Line 1: foo() is called and assigns 1 to global x and prints it as no other local x are declared in it.

Line 2: printx is called with global x

Line 3: boo() is called and it has local x declared, so latest scope of x is that inside boo()'s. So boo() calls assignx and sets its local x to 2 and then calls printx with this same x for printing it.

Line 4: finally printx is called with global x's scope which prints 1 as it was set in line 2

Thanking you...