CSCI 3136 Principles of Programming Languages Control Flow - 2 - - PowerPoint PPT Presentation

CSCI 3136 Principles of Programming Languages

Control Flow - 2

Summer 2013 Faculty of Computer Science Dalhousie University

1 / 40

Short-Circuit Evaluation of Boolean Expressions

• (and a b): If a is false, b has no effect on the value of the whole

expression.

• (or a b): If a is true, b has no effect on the value of the whole

expression.

2 / 40

Short-Circuit Evaluation of Boolean Expressions

• (and a b): If a is false, b has no effect on the value of the whole

expression.

• (or a b): If a is true, b has no effect on the value of the whole

expression. Short-circuit evaluation

• If the value of the expression does not depend on b, the evaluation
• f b is skipped.

This is a useful optimization. If the evaluation of b has side-effects, however, the meaning of the code may be changed.

3 / 40

Short-Circuit Evaluation of Boolean Expressions

• (and a b): If a is false, b has no effect on the value of the whole

expression.

• (or a b): If a is true, b has no effect on the value of the whole

expression. Short-circuit evaluation

• If the value of the expression does not depend on b, the evaluation
• f b is skipped.

This is a useful optimization. If the evaluation of b has side-effects, however, the meaning of the code may be changed. Some languages provide both regular and short-circuit versions of Boolean operators. Ada

• and vs and then
• or vs or else

4 / 40

Short-Circuit Evaluation: Examples

• C: (C short-circuits && and || operators)

while( p != NULL && p->e != val ) p = p->next;

5 / 40

Short-Circuit Evaluation: Examples

• C: (C short-circuits && and || operators)

while( p != NULL && p->e != val ) p = p->next;

• Pascal:(Pascal does not short-circuit and and or operators )

p := my list; while (p <> nil) and (p^ .key <> val) do p := p^ .next;

6 / 40

Short-Circuit Evaluation: Examples

• C: (C short-circuits && and || operators)

while( p != NULL && p->e != val ) p = p->next;

• Pascal:(Pascal does not short-circuit and and or operators )

p := my list; while (p <> nil) and (p^ .key <> val) do p := p^ .next;

• Perl:
• pen( F, "file" ) or die;

7 / 40

Short-Circuit Evaluation: Examples

• C: (C short-circuits && and || operators)

while( p != NULL && p->e != val ) p = p->next;

• Pascal:(Pascal does not short-circuit and and or operators )

p := my list; while (p <> nil) and (p^ .key <> val) do p := p^ .next;

• Perl:
• pen( F, "file" ) or die;
• Replacing if in Perl or shell scripts:

if( x > max ) then max = x; becomes ( x > max ) and max = x;

8 / 40

Sequencing

In imperative programming languages, sequencing comes naturally, without a need for special syntax to support it. Mixed imperative/functional languages (LISP, Scheme, ...) often provide special constructs for sequencing. Issue: What’s the value of a sequence of expressions/statements?

• The value of the last subexpression (most common)

C: a = 4, b = 5 LISP: (progn (setq a 4) (setq b 5))

• The value of the first subexpression

LISP: (prog1 (setq a 4) (setq b 5))

• The value of the second subexpression (supported in LISP)

LISP: (prog2 (setq a 4) (setq b 5) (setq c 6))

9 / 40

Sequencing

In imperative programming languages, sequencing comes naturally, without a need for special syntax to support it. Mixed imperative/functional languages (LISP, Scheme, ...) often provide special constructs for sequencing. Issue: What’s the value of a sequence of expressions/statements?

• The value of the last subexpression (most common)

C: a = 4, b = 5 = ⇒ 5 LISP: (progn (setq a 4) (setq b 5))

• The value of the first subexpression

LISP: (prog1 (setq a 4) (setq b 5))

• The value of the second subexpression (supported in LISP)

LISP: (prog2 (setq a 4) (setq b 5) (setq c 6))

10 / 40

Sequencing

In imperative programming languages, sequencing comes naturally, without a need for special syntax to support it. Mixed imperative/functional languages (LISP, Scheme, ...) often provide special constructs for sequencing. Issue: What’s the value of a sequence of expressions/statements?

• The value of the last subexpression (most common)

C: a = 4, b = 5 = ⇒ 5 LISP: (progn (setq a 4) (setq b 5)) = ⇒ 5

• The value of the first subexpression

LISP: (prog1 (setq a 4) (setq b 5))

• The value of the second subexpression (supported in LISP)

LISP: (prog2 (setq a 4) (setq b 5) (setq c 6))

11 / 40

Sequencing

In imperative programming languages, sequencing comes naturally, without a need for special syntax to support it. Mixed imperative/functional languages (LISP, Scheme, ...) often provide special constructs for sequencing. Issue: What’s the value of a sequence of expressions/statements?

• The value of the last subexpression (most common)

C: a = 4, b = 5 = ⇒ 5 LISP: (progn (setq a 4) (setq b 5)) = ⇒ 5

• The value of the first subexpression

LISP: (prog1 (setq a 4) (setq b 5)) = ⇒ 4

• The value of the second subexpression (supported in LISP)

LISP: (prog2 (setq a 4) (setq b 5) (setq c 6))

12 / 40

Sequencing

In imperative programming languages, sequencing comes naturally, without a need for special syntax to support it. Mixed imperative/functional languages (LISP, Scheme, ...) often provide special constructs for sequencing. Issue: What’s the value of a sequence of expressions/statements?

• The value of the last subexpression (most common)

C: a = 4, b = 5 = ⇒ 5 LISP: (progn (setq a 4) (setq b 5)) = ⇒ 5

• The value of the first subexpression

LISP: (prog1 (setq a 4) (setq b 5)) = ⇒ 4

• The value of the second subexpression (supported in LISP)

LISP: (prog2 (setq a 4) (setq b 5) (setq c 6)) = ⇒ 5

13 / 40

Goto and Alternatives

Use of goto is bad programming practice if the same effect can be achieved using different constructs. Sometimes, however, it is unavoidable:

• Break out of a loop
• Break out of a subroutine
• Break out of a deeply nested context

Many languages provide alternatives:

• One-and-a-half loop
• return statement
• Structured exception handling

14 / 40

Selection (Alternation)

• Standard if-then-else statement

if ... then ... else ...

• Multi-way if-then-else statement

if ... then ... elsif ... then ... elsif ... then ... ... else ...

• Switch statement

switch ... of case ...: ... case ...: ... ...

15 / 40

Switch Statements

Switch statements are a special case of if/then/elsif/else. Principal motivation: Generate more efficient code. Compiler can use different methods to generate efficient code:

• Sequential testing
• Binary search
• Hash table
• Jump table

16 / 40

Implementation of Switch Statements (1)

i := ... (* expression *) IF i = 1 THEN clause A ELSIF i IN 2, 7 THEN clause B ELSIF i IN 3..5 THEN clause C ELSIF i = 10 THEN clause D ELSE clause E END —————————— CASE i OF 1: clause A | 2, 7: clause B | 3..5: clause C | 10: clause D ELSE clause E END r1:= ...

• - expression

if r1 = 1 goto L1 clause A goto L6 L1: if r1 = 2 goto L2 if r1 = 7 goto L3 L2: clause B goto L6 L3: if r1 < 3 goto L4 if r1 > 5 goto L4 clause C goto L6 L4: if r1 = 10 goto L5 clause D goto L6 L5: clause E L6: ...

17 / 40

Implementation of Switch Statements (1)

i := ... (* expression *) IF i = 1 THEN clause A ELSIF i IN 2, 7 THEN clause B ELSIF i IN 3..5 THEN clause C ELSIF i = 10 THEN clause D ELSE clause E END —————————— CASE i OF 1: clause A | 2, 7: clause B | 3..5: clause C | 10: clause D ELSE clause E END r1:= ...

• - expression

if r1 = 1 goto L1 clause A goto L6 L1: if r1 = 2 goto L2 if r1 = 7 goto L3 L2: clause B goto L6 L3: if r1 < 3 goto L4 if r1 > 5 goto L4 clause C goto L6 L4: if r1 = 10 goto L5 clause D goto L6 L5: clause E L6: ...

18 / 40

Implementation of Switch Statements (2)

i := ... (* expression *) IF i = 1 THEN clause A ELSIF i IN 2, 7 THEN clause B ELSIF i IN 3..5 THEN clause C ELSIF i = 10 THEN clause D ELSE clause E END —————————— CASE i OF 1: clause A | 2, 7: clause B | 3..5: clause C | 10: clause D ELSE clause E END Jump table T: &L1 &L2 &L3 &L3 &L3 &L5 &L2 &L5 &L5 &L4 L6: r1:= ...--expression if r1 < 1 goto L5 if r1 > 10 goto L5 r1 := r1 -1 r2 := T[r1] goto *r2 L1: clause A goto L7 L2: clause B goto L7 L3: clause C goto L7 L4: clause D goto L7 L5: clause E goto L7 L7: ...

19 / 40

Implementation of Switch Statements (2)

i := ... (* expression *) IF i = 1 THEN clause A ELSIF i IN 2, 7 THEN clause B ELSIF i IN 3..5 THEN clause C ELSIF i = 10 THEN clause D ELSE clause E END —————————— CASE i OF 1: clause A | 2, 7: clause B | 3..5: clause C | 10: clause D ELSE clause E END Jump table T: &L1 &L2 &L3 &L3 &L3 &L5 &L2 &L5 &L5 &L4 L6: r1:= ...--expression if r1 < 1 goto L5 if r1 > 10 goto L5 r1 := r1 -1 r2 := T[r1] goto *r2 L1: clause A goto L7 L2: clause B goto L7 L3: clause C goto L7 L4: clause D goto L7 L5: clause E goto L7 L7: ...

20 / 40

Implementation of Switch Statements (3)

Jump table + Fast: one table lookup to find the right branch − Potentially large table: one entry per possible value Linear search − Potentially slow + No storage overhead Hash table + Fast: one hash table access to find the right branch − More complicated − Elements in a range need to be stored individually ⇒ again, possibly large table Binary search ± Fast (but slower than table lookup) + No storage overhead No single implementation is best in all circumstances. Compilers often use different strategies based on the specific code.

21 / 40

Iteration

Enumeration-controlled loops

• Example: for-loop
• One iteration per element in a finite set.
• The number of iterations is known in advance.

Logically controlled loops

• Example: while and repeat loops
• Executed until a Boolean condition changes.
• The number of iterations is not known in advance.

22 / 40

Iteration

Enumeration-controlled loops

• Example: for-loop
• One iteration per element in a finite set.
• The number of iterations is known in advance.

Logically controlled loops

• Example: while and repeat loops
• Executed until a Boolean condition changes.
• The number of iterations is not known in advance.

Some languages do not have loop constructs (e.g., Scheme). They use tail recursion instead.

23 / 40

Logically Controlled Loops

• Pre-loop test

while ... do ...

• Post-loop test

repeat ... until ... do ... while ...

• Mid-loop test or ”one-and-a-half loop”

loop ... when ... exit ... when ... exit ... end

24 / 40

Logically Controlled Loops

• Pre-loop test

while ... do ...

• Post-loop test

repeat ... until ... do ... while ... Loop is executed at least once.

• Mid-loop test or ”one-and-a-half loop”

loop ... when ... exit ... when ... exit ... end

25 / 40

Implementation of Iterations (1)

WHILE cond do statements END ——————————— L1: r1 = evaluate cond if not r1 goto L2 statements goto L1 L2: ...

26 / 40

Implementation of Iterations (1)

WHILE cond do statements END ——————————— L1: r1 = evaluate cond if not r1 goto L2 statements goto L1 L2: ... for( init; cond; step ) { statements } ———————————

27 / 40

Implementation of Iterations (1)

WHILE cond do statements END ——————————— L1: r1 = evaluate cond if not r1 goto L2 statements goto L1 L2: ... for( init; cond; step ) { statements } ——————————— init L1: r1 = evaluate cond if not r1 goto L2 statements step goto L1 L2: ...

28 / 40

Implementation of Iterations (2)

Potentially much more efficient:

FOR i = start TO end BY step DO statements END

If modifying the loop variable inside the loop is allowed:

——————————— r1 := start r2 := end r3 := step L1: if r1 > r2 goto L2 statements r1 := r1 + r3 goto L1 L2: ...

If modifying the loop variable inside the loop is not allowed:

——————————— r1 := [(end - start)/step]+1 L1: if not r1 goto L2 statements decrement r1 goto L1 L2: ...

29 / 40

Break and Continue

”Break” statement (”last” in Perl)

• Exits the nearest enclosing for, do, while or switch statement.

”Continue” statement (”next” in Perl)

• Skips the rest of the current iteration.

The loop may have a finally part, which is always executed no matter whether the iteration is executed normally or terminated using a continue or break statement.

30 / 40

Recursion

• Recursion replaces iteration in functional programming paradigm.

31 / 40

Recursion

• Recursion replaces iteration in functional programming paradigm.
• For example:

while( condition ) { S1; S2; ... }

32 / 40

Recursion

• Recursion replaces iteration in functional programming paradigm.
• For example:

while( condition ) { S1; S2; ... }

• We can use:

procedure P() { if( condition ) { S1; S2; ...; P(); } } Iteration is a strictly imperative feature: it relies on updating the iterator variable.

33 / 40

Implementation of Recursion

• A naive implementation of recursion is often less efficient than a

naive implementation of iteration.

34 / 40

Implementation of Recursion

• A naive implementation of recursion is often less efficient than a

naive implementation of iteration.

• An optimizing compiler often converts recursion into iteration when

possible:

− Tail recursion: There is no work to be done after the recursive call. − More general recursion: Work to be done after the recursive call may be passed to the recursive call as a continuation.

35 / 40

Tail-recursion

• There is no work to be done after the recursive call.
• Tail-recursive calls are often optimized into an iterative form,

especially in functional languages.

• An optimized tail-recursive call does not allocate additional stack

space and does not have function call overhead.

36 / 40

Tail-recursion (Example in C)

• Consider this recursive definition of the factorial function in C:

factorial(n) { if (n == 0) return 1; return n * factorial(n - 1); }

37 / 40

Tail-recursion (Example in C)

• Consider this recursive definition of the factorial function in C:

factorial(n) { if (n == 0) return 1; return n * factorial(n - 1); }

• tail-recursive version:

factorial1(n, accumulator) { if (n == 0) return accumulator; return factorial1(n - 1, n * accumulator); } factorial(n) { return factorial1(n, 1); }

38 / 40

Tail-recursion (Example in Scheme)

• Consider this Scheme function to sum the elements of a list:

(define (sum lis) (if (null? lis) (+ (car lis) (sum (cdr lis)))))

39 / 40

Tail-recursion (Example in Scheme)

• Consider this Scheme function to sum the elements of a list:

(define (sum lis) (if (null? lis) (+ (car lis) (sum (cdr lis)))))

• tail-recursive version:

(define (sum lis) (letrec ((helper (lambda (s lis) (if (null? lis) s (helper (+ s (car lis)) (cdr lis)))))) (helper 0 lis)))

40 / 40