Data structure routine -- Expression evaluate (using stack structure), routine evaluate

Data structure routine -- Expression evaluate (using stack structure), routine evaluate

This article focuses on the basic series of network courses on data structures (3): Application 1-expression evaluation of 5th class stacks in stacks and queues.

For example, you enter a valid mathematical expression that contains "+", "-", "*", "/", positive integers, and parentheses to calculate the operation result of the expression.

Answer:

# Include <stdio. h> # include <stdlib. h> # define MaxOp 100 # define MaxSize 100 struct // sets the operator priority {char ch; // operator int pri; // priority} lpri [] = {'= ', 0}, {'(', 1}, {'*', 5}, {'/', 5}, {'+', 3 },{'-', 3}, {')', 6 }}, rpri [] = {'=', 0}, {'(', 6 },{'*', 4}, {'/', 4}, {'+', 2}, {'-', 2}, {')', 1 }}; int leftpri (char op) // evaluate the op priority of the Left operator {int I; for (I = 0; I <MaxOp; I ++) if (lpri [I]. ch = op) return lpri [I]. pri;} int rightpri (char op) // evaluate the op priority of the right operator {int I; for (I = 0; I <MaxOp; I ++) if (rpri [I]. ch = op) return rpri [I]. pri;} bool InOp (char ch) // determines if ch is an operator {if (ch = '(' | ch = ') '| ch =' + '| ch ='-'| ch =' * '| ch ='/') return true; else return false;} int Precede (char op1, char op2) // comparison result of op1 and op2 operator priority {if (leftpri (op1) = rightpri (op2 )) return 0; else if (leftpri (op1) <rightpri (op2) return-1; else return 1;} void trans (char * exp, char postexp []) // calculate Convert the expression exp to the suffix expression postexp {struct {char data [MaxSize]; // store the int top operator; // Stack pointer} op; // define the operator stack int I = 0; // I is the subscript op of postexp. top =-1; op. top ++; // Add '=' to the stack op. data [op. top] = '; while (* exp! = '\ 0') // The loop {if (! InOp (* exp) // when it is a numeric character {while (* exp> = '0' & * exp <= '9 ') // It is determined to be a number {postexp [I ++] = * exp; exp ++;} postexp [I ++] = '#'; // use # To identify the end of a numeric string} else // switch (Precede (op. data [op. top], * exp) {case-1: // low priority of the stack top operator: Stack op. top ++; op. data [op. top] = * exp; exp ++; // Continue scanning other characters break; case 0: // only parentheses meet this condition op. top --; // run (return stack exp ++; // Continue scanning other characters break; case 1: // return the stack and output it to postexp. postexp [I ++] = op. data [op. top]; op. top --; break ;}} // While (* exp! = '\ 0') while (op. data [op. top]! = ') // At this time, the exp scan is completed and the stack is '=' until {postexp [I ++] = op. data [op. top]; op. top --;} postexp [I] = '\ 0'; // Add an end identifier to the postexp expression} float compvalue (char exp []) // calculate the value of the suffix expression {struct {float data [MaxSize]; // store the value int top; // Stack pointer} st; // define the value stack float d; char ch; int t = 0; // t is the subscript st of exp. top =-1; ch = exp [t]; t ++; while (ch! = '\ 0') // The loop {switch (ch) {case' + ': st. data [st. top-1] = st. data [st. top-1] + st. data [st. top]; st. top --; break; case '-': st. data [st. top-1] = st. data [st. top-1]-st. data [st. top]; st. top --; break; case '*': st. data [st. top-1] = st. data [st. top-1] * st. data [st. top]; st. top --; break; case '/': if (st. data [st. top]! = 0) st. data [st. top-1] = st. data [st. top-1]/st. data [st. top]; else {printf ("\ n \ t Division Error! \ N "); exit (0); // exit unexpectedly} st. top --; break; default: d = 0; // convert a number to a value and store it in d. while (ch> = '0' & ch <= '9 ') // It is a numeric character {d = 10 * d + ch-'0'; ch = exp [t]; t ++;} st. top ++; st. data [st. top] = d;} ch = exp [t]; t ++;} return st. data [st. top];} int main () {char exp [] = "(56-20)/(4 + 2 )"; // you can change exp to the keyboard input char postexp [MaxSize]; trans (exp, postexp); printf ("infix expression: % s \ n", exp ); printf ("suffix expression: % s \ n", postexp); printf ("expression value: % g \ n", compvalue (postexp); return 0 ;}

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