[SOLVED] CS 431 Programming language concepts Homework 1

Homework 1
September 11, 2018
For this homework, you will write BNF grammars. You do not need to
resolve ambiguity such as operator precedence or associativity. Your grammars
do not consider white spaces.
Assume that there exists the non-terminal <num> that represents numbers.
1. Dene a non-terminal <var> that represents the variables a, b, c, x, y, and
z.
2. Dene a non-terminal <aop> that represents binary arithmetic operators
+, -, *, /, ^.
3. Dene a non-terminal <aexp> that represents arithmetic expressions. The
grammar should accept expressions such as: ((a – 5) / c + 1) * 2,
2 * (x + 3) * y + z / a, and 2^44 + b. Please utilize the non-terminals
such as <var>, <num>, and <aop>.
4. Dene a non-terminal <bop> that represents binary boolean operators &&
and jj. Note that 0jj0 is quoted since j is also used as separator for the
grammar.
5. Dene a non-terminal <cop> that represents binary comparison operators
<,<=,==,!=,>=,>
6. Dene a non-terminal <bexp> that represent logical expressions (i.e. boolean
expressions). The grammar should accept expressions such as:
!(a/2 <= 10 && 5 != c)
a < c || b > 0
Note that logical expressions include negation operator ! and the boolean
constant true and false. You should utilize the non-terminals <aexp>,
<bop>, <cop>.
7. Dene a non-terminal <Stmt> that represents statements that include as-
signments, while loops, and if-statements (with optional else-part.) The
statements may include arithmetic, comparison, and boolean expressions.
Dene a non-terminal <Stmts> that represents zero or more statements.
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Note that <Stmt> and <Stmts> are mutually dependent.
You may use the special non-terminal <empty> to represent nothing.
You should utilize the non-terminals <var>, aexp, <bexp>.
Here is program that can be represented by the non-terminal <Stmts>.
b = true;
y = 2;
while(!b && y < x) {
if ((y – (y/x)*x) == 0) {
b = false;
}
}
if(b) {
z = y/x;
}
else {
z = 1;
}
Here is another program that can be represented by <Stmts>.
x = 1;
y = x;
if (a > x) {
while (x <= a) {
x = x + 1;
y = y * x;
}
}
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