BBM 201 DATA STRUCTURES

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1 BBM 201 DATA STRUCTURES Lecture 6: EVALUATION of EXPRESSIONS 2017 Fall

2 Evaluation of Expressions Compilers use stacks for the arithmetic and logical expressions. Example: x=a/b-c+d*e-a*c If a=4, b=c=2, d=e=3 what is x? ((4/2)-2)+(3*3)-(4*2), ( / and * have a priority) There may be also parenthesis, such as: a/(b-c)+d*(e-a)*c How does the compiler solve this problem?

3 Infix, prefix, postfix Normally, we use infix notation for the arithmetic expressions: Infix notation: a+b However, there is also prefix and postfix notation: Prefix notation: +ab Postfix notation: ab+ Infix : 2+3*4 Postfix: 234*+ Prefix: +2*34

4 Prefix

5 Postfix

6 How to convert infix to prefix? Move each operator to the left of the operands: Operand order does not change!

7 How to convert infix to postfix? Move each operator to the right of the operands: Operand order still does not change!

8 Example - 1 Infix: (a+b)*c-d/e Postfix:??? Prefix:???

9 Example 1 (solution) Infix: (a+b)*c-d/e Postfix: ab+c*de/- Prefix: -*+abc/de

10 Example - 2 Infix: a/b-c+d*e-a*c Postfix:??? Prefix:???

11 Example 2 (solution) Infix: a/b-c+d*e-a*c Postfix: ab/c-de*+ac*- Prefix:-+-/abc*de*ac

12 Example 3 Infix: (a/(b-c+d))*(e-a)*c Postfix:??? Prefix:???

13 Example 3 (solution) Infix: (a/(b-c+d))*(e-a)*c Postfix: abc-d+/ea-*c* Prefix: **/a+-bcd-eac

14 Expressions Infix Postfix Prefix Notes A * B + C / D A B * C D / + + * A B / C D A * (B + C) / D A B C + * D / / * A + B C D A * (B + C / D) A B C D / + * * A + B / C D multiply A and B, divide C by D, add the results add B and C, multiply by A, divide by D divide C by D, add B, multiply by A

15 Infix, prefix, postfix

16 Evaluation of a postfix expression Token Stack [0] [1] [2] / 4/ / (4/2) (4/2) ((4/2)-2) * ((4/2)-2) 3*3 1 + ((4/2)-2)+(3*3) 0 4 ((4/2)-2)+(3*3) ((4/2)-2)+(3*3) * ((4/2)-2)+(3*3) 4*2 1 - ((4/2)-2)+(3*3)-(4*2) 0 Top

17 Why postfix? For the infix expressions we have two problems: Parenthesis Operation precedence Example: ((4/2)-2)+(3*3)-(4*2) (infix) 42/2-33*+42*- (postfix)

18 Operator PRECEDENCE Operators Associativity Type ! (type) right to left unary * / % left to right multiplicative + - left to right additive < <= > >= left to right relational ==!= left to right equality && left to right logical AND left to right logical OR?: right to left conditional = += -= *= /= %= right to left assignment, left to right comma Fig Operator precedence and associativity. Parentheses are used to override precedence.

19 EVALUATION OF INFIX OPERATIONS (fully Parenthesized) 1. Read one input character 2. Actions at the end of each input Opening brackets (2.1) Push into stack and then Go to step (1) Number (2.2) Push into stack and then Go to step (1) Operator (2.3) Push into stack and then Go to step (1) Closing brackets (2.4) Pop from character stack New line character (2.4.1) if it is opening bracket, then discard it, Go to step (1) (2.4.2) Pop is used four times The first popped element is assigned to op2 The second popped element is assigned to op The third popped element is assigned to op1 The fourth popped element is the remaining opening bracket, which can be discarded Evaluate op1 op op2 Convert the result into character and push into the stack Go to step (2.4.) (2.5) Pop from stack and print the answer STOP

20 Input Symbol (((2 * 5) - (1 * 2)) / (11-9)) ( ( ( ( ( ( ( ( ( 2 ( ( ( 2 * ( ( ( 2 * 5 ( ( ( 2 * 5 Stack (from bottom to top) Operation ) ( ( 10 2 * 5 = 10 and push - ( ( 10 - ( ( ( 10 - ( 1 ( ( 10 - ( 1 * ( ( 10 - ( 1 * 2 ( ( 10 - ( 1 * 2 ) ( ( * 2 = 2 & Push ) ( = 8 & Push / ( 8 / ( ( 8 / ( 11 ( 8 / ( 11 - ( 8 / ( 11-9 ( 8 / ( 11-9 ) ( 8 / = 2 & Push ) 4 8 / 2 = 4 & Push New line Empty Pop & Print

21 EVALUATION OF INFIX OPERATIONS (Not fully Parenthesized) 1. Read an input character 2. Actions that will be performed at the end of each input operator with stack into op2 Opening parentheses (2.1) Push it into character stack and then Go to step 1 Number (2.2) Push into integer stack, Go to step 1 Operator (2.3) Do the comparative priority check Closing par. (2.4) Pop from the character stack (2.3.1) if the character stack's top contains an equal or higher priority, Then pop it into op Pop a number from integer Pop another number from integer stack into op1 Calculate op1 op op2 and Push the result into the integer stack (2.4.1) if it is an opening parentheses, then discard it and Go to step 1 it to op1 integer stack New line character (2.4.2) To op, assign the popped element Pop a number from integer stack and assign it op2 Pop another number from integer stack and assign Calculate op1 op op2 and push the result into the Convert into character and push into stack Go to the step 2.4 (2.5) Print the result after popping from the stack STOP

22 (2*5-1*2)/(11-9) Input Symbol Character Stack (from bottom to top) Integer Stack (from bottom to top) Operation performed ( ( 2 ( 2 * ( * Push as * has higher priority 5 ( * ( * Since '-' has less priority, we do 2 * 5 = 10 ( - 10 We push 10 and then push '-' 1 ( * ( - * 10 1 Push * as it has higher priority 2 ( - * ) ( Perform 1 * 2 = 2 and push it ( 8 Pop - and 10-2 = 8 and push, Pop ( / / 8 ( / ( 8 11 / ( / ( / ( ) / 8 2 Perform 11-9 = 2 and push it 4 Perform 8 / 2 = 4 and push it New line 4 Print the output, which is 4

23 Evaluation of a prefix operation Input: / - * 2 5 * Output: 4 Data structure requirement: a character stack and an integer stack 1. Read one character input at a time and keep pushing it into the character stack until the new line character is reached 2. Perform pop from the character stack. If the stack is empty, go to step (3) Number (2.1) Push into the integer stack and then go to step (2) Operator (2.2) Assign the operator to op Pop a number from integer stack and assign it to op1 Pop another number from integer stack and assign it to op2 Calculate op1 op op2 and push the output into the integer stack. Go to step (2) 3. Pop the result from the integer stack and display the result

24 / - * 2 5 * / / - /- * / - * 2 / - * 2 5 / - * 2 5 * / - * 2 5 * 1 / - * 2 5 * 1 2 / - * 2 5 * / - * 2 5 * / - * 2 5 * / - * 2 5 * \n / - * 2 5 * / - * 2 5 * / - * 2 5 * = 2 / - * 2 5 * / - * 2 5 * / - * * 2 = 2 / - * / - * / * 2 = 10 / = 8 Stack is empty 4 8 / 2 = 4 Stack is empty Print 4

25 POSTFIX Compilers typically use a parenthesis-free notation (postfix expression). The expression is evaluated from the left to right using a stack: when encountering an operand: push it when encountering an operator: pop two operands, evaluate the result and push it.

26 62/3-42*+

27 How to evaluate a postfix evaluation? typedef enum {left_parent,right_parent,add,subtract,multiply,divide,eos,operand} precedence; char expr[]= "422-3+/34-*2*";; precedence get_token(char* symbol, int* n){ } *symbol=expr[(*n)++]; switch(*symbol){ case ( : return left_parent; case ) : return right_parent; case + : return add; case - : return subtract; case / : return divide; case * : return multiply; case \0 : return eos; default: return operand; }

28 How to evaluate a postfix evaluation? float eval(void){ char symbol; precedence token; float op1, op2; int n=0; int top=-1; token = get_token(&symbol, &n); //take a token } while(token=eos){ //end of string? if(token==operand) push(&top,symbol- 0 ); else{ op2=pop(&top); op1=pop(&top); switch(token){ case add: push(&top,op1+op2); case subtract: push(&top,op1-op2); case multiply: push(&top,op1*op2); case divide: push(&top,op1/op2); } } token=get_token(&symbol,&n); } return pop(&top);

29 CONVERT an INFIX to POSTFIX a+b*c a*(b+c)*d

30 How to convert infix to postfix? char expr[]=" (4/(2-2+3))*(3-4)*2";; static int stack_pre[]={0,19,12,12,13,13,13,0};; static int pre[]={20,19,12,12,13,13,13,0};; void postfix(void){ char symbol; precedence token; int n=0; int top=0; stack[0]=eos; for(token=get_token(&symbol,&n);token!=eos;token=get_token(&symbol,&n)){ if(token==operand) printf("%c", symbol); else if(token==right_parent) while(stack[top]!=left_parent) print_token(pop(&top)); pop(&top); } else{ while(stack_pre[stack[top]]>=pre[token]) print_token(pop(&top)); push(&top, token); } } while((token=pop(&top))!=eos) print_token(token) }

31 Exercise to do at home: 1. Write the code that converts infix to prefix. 2. Write the code that evaluates a prefix expression.

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