CS471, Programming Language Structure I
Fall, 1998

Examination 1

Closed Book. Answer all the questions. The total points for each question or part of a question follows it in parentheses, thus: (12)

1

Java has both a compiler (which produces a "byte-code" version of the program) and an interpreter that interprets the byte codes. Draw a diagram that shows how a Java program produces output results. You should include the following features: compiler, interpreter, run-time library, source code, byte code. (16)

2

Below is an Ada loop construct. Executing the exit command causes the loop to terminate immediately. Transform the loop into an equivalent form that uses the standard structured programming while loop. You may use C or Pascal syntax, or you may draw the flow chart as your answer. (18)

x := 1;
loop
y := x + 1;
if y > 10 then exit;
x := y + 2;
end loop;

3

Look at the statement sequence below, written in a strongly-typed language. Explain how each statement is handled at compile and/or run time if:

the language is statically typed (i.e. all variables must have a type declaration). (12)

the language is dynamically typed (i.e. the declarations can be omitted) (12)

x, y : integer;
x = 1;
y = x + 1;
x = "abcd";
y = x + 1;

4

In the program below, written in C syntax, what is the value of g in the body of f1 with:

static scoping (8), and

dynamic scoping? (8)

By changing the order of function calls, show how g in the body of f1 could have the value 3 with dynamic scoping. (4)

int g = 1;
void f1();
void f2() {
f1();
}

void f3() {
int g = 2;
f2();
}

void f1() {
/* what is the value of g here? */
}

int main() {
int g = 3;
f3();
}

5

Below is a grammar for a portion of Pascal. Show, by drawing a parse tree, that the string:

begin S1; begin S2; S3 end; begin S1; S3 end end

can be generated by the grammar. (16) Then alter the grammar, by making a simple change, so that the semi-colon becomes a statement terminator (as in C), not a separator, as it is in Pascal. (6)

<compound-stmt> ::= begin <stmt-seq> end
<stmt-seq> ::= <stmt> | <stmt> ; <stmt-seq>
<stmt> ::= <simple-stmt> | <compound-stmt>
<simple-stmt> ::= S1 | S2 | S3