# Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 DCO 20105 Data structures and algorithms  Lecture 12: Stack and Expression Evaluation  Stack basic.

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Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 DCO 20105 Data structures and algorithms  Lecture 12: Stack and Expression Evaluation  Stack basic operations of a stack its implementation using a vector applications of stack  Expression Evaluation Infix, Prefix, and Postfix expressions Implementation of evaluating a postfix expression -- By Rossella Lau

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Stack  Stack is an ordered list of items, ordered in the input sequence  Basically, items  are inserted and deleted at one end, called the top of the stack  are last in first out (LIFO)  An example: The original Take an item Add an item A B C A B A B D A B D E C

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Operations of a stack  Basic:  push() – to add an item  pop() – to remove an item  Assistance:  size() – returns the number of items in a stack  empty() – to check if a stack is empty  top() – returns the first element of a stack; note that in a queue, it is called front()

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Exercises  Ford: 7: 13,14; 8:11

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Implementation of a stack using a vector // Stack.h template class Stack { private: vector stack; public: void push(T const & item) {stack.push_back(item);} T & pop() {T t(top()); stack.pop_back(); return t;} size_t size() const { return stack.size();} bool empty() const { return stack.empty();} T const & top() const { return stack.back();} };

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Applications of stack  Nested parentheses check  Expression evaluation  The underlying structure for recursion  Function calls (and its associated variables), e.g., : main check init main check init main check push main check push pop …

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 A stack application: nested parentheses check  Consider the parentheses in mathematical expressions 1. Equal number of right and left parentheses 2. Every right parenthesis is preceded by a matching parenthesis  Wrong expressions: ((A+B) violate condition 1 )A+B(-Cviolate condition 2  Solution: nesting depth parenthesis count

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Use of nesting depth and parenthesis count  Nesting depth: left parenthesis: open a scope right parenthesis: close a scope nesting depth at a particular point is the number of scopes that have been opened but not closed  Parenthesis count: at a particular point as the number of left parentheses minus the number of right parentheses if the count is not negative, it is the nesting depth  Check parentheses use by checking if parenthesis count: is greater than or equal to 0 at any point is equal to 0 at the end of the expression of course, checking should include the parenthesis type: {}, [],()

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Use of a stack to check  For each left parenthesis, do a push  For each right parenthesis, do a pop and make sure the item popped is equal to the right parenthesis (same type)  If the stack is empty when doing a pop  ERROR  If the stack is not empty when the expression ends  ERROR  Examples of checking invalid expressions: ( (( ((A+B) !empty(s) )A+B(-C Cannot pop ( [(A+B]) [ ( Type mismatch ( [ (

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Algorithm to check pairs of parentheses bool check(string expression){ valid = true stack brackets while (expression not end){ read a symbol (symb) if symb is a left bracket brackets.push(symb) if symb is a right bracket{ if brackets.empty() valid=false if brackets.front() not match symb valid = false brackets.pop() } } if !brackets.empty() valid = false return valid} { {x+(y-[a+b])*c-(d+e)}/(h-(j-(k-[l-n]))) ( [ {x+(y-[ [ ( { { … a+b]) ( {{ ( { … *c-( { … d+e) { {…}{…} ( ( ( …/(h-(j-(k-[ [ ( ( ( [ …l-n])))

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Expression evaluation  An expression is a series of operators and operands  Operators: +, -, *, /, %, \$ (exponentiation)  Operands: the values going to be operated Three representations of an expression:  Infix: The usual form, operator is between operands  Prefix: the operator is in front of the operand(s)  Postfix: the operand(s) is(are) in front of the operator

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Examples of Prefix Expressions Infix A+B A+B-C (A+B)*(C-D) A\$B*C-D+E/F/(G+H) ((A+B)*C-(D-E))\$(F+G) A-B/(C*D\$E) Prefix +AB -+ABC *+AB-CD +-*\$ABCD//EF+GH \$-*+ABC-DE+FG -A/B*C\$DE

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Examples of Postfix Expressions Infix A+B A+B-C (A+B)*(C-D) A\$B*C-D+E/F/(G+H) ((A+B)*C-(D-E))\$(F+G) A-B/(C*D\$E) Postfix AB+ AB+C- AB+CD-* AB\$C*D-EF/GH+/+ AB+C*DE--FG+\$ ABCDE\$*/-

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Evaluating a postfix expression  The advantage is no parentheses, i.e., less complication  Since the operator comes later, operands can be pushed to the stack first and once an operator is encountered, the operands are popped for the calculation  The algorithm (one digit operands) : /* for one digit operands and no space in the expression */ int eval(string const & expr) { Stack operands; for (size_t i=0; i { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3591651/slides/slide_15.jpg", "name": "Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Evaluating a postfix expression  The advantage is no parentheses, i.e., less complication  Since the operator comes later, operands can be pushed to the stack first and once an operator is encountered, the operands are popped for the calculation  The algorithm (one digit operands) : /* for one digit operands and no space in the expression */ int eval(string const & expr) { Stack operands; for (size_t i=0; i

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 The algorithm for more than one digit double eval(string const & expr) { Stack operands; char c, number[MAX]; size_t i, j=0; double num; for (i=0; i { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3591651/slides/slide_16.jpg", "name": "Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 The algorithm for more than one digit double eval(string const & expr) { Stack operands; char c, number[MAX]; size_t i, j=0; double num; for (i=0; i

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 cal(char, double, double) double cal(char oper, double op1,double op2) { switch (oper) { // “operator” is a key word case '+': return op1 + op2; case '-': return op1 - op2; case '*': return op1 * op2; case '/': return op1 / op2; case '\$': return pow(op1,op2); default: // should not happen exit(EXIT_FAILURE); }

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 An example of postfix expression evaluation  4 5 7 + *  scan 457: push(4), push(5), push(7)  scan +: op2=7, pop(), op1=5, pop(), push(op1+op2)  scan *: op2=12, pop(), op1=4, pop(), push(op1*op2)  expression end ==> return 48!

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Summary  A stack is one of the most basic data structures in the study.  It only has one end for insert and delete.  An item in a stack is LIFO while it is FIFO in a queue  Stack is used in many hidden areas of computer systems  One popular important application is expression evaluation  An expression can be in three forms: infix, prefix, and postfix  It is common to evaluate an expression by using a postfix format

Rossella Lau Lecture 12, DCO20105, Semester A,2005-6 Reference  Ford: 7  Example programs: evaluateIntExpression.cpp, evaluateDoubleExpression.cpp  STL online references  http://www.sgi.com/tech/stl http://www.sgi.com/tech/stl  http://www.cppreference.com/ http://www.cppreference.com/ -- END --

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