Expression solving problems (using stacks)

problem description ]

Expression solving is one of the basic problems of implementing programming , and it is also a typical example of the application of stack . Design a program to evaluate an expression using operator precedence .

[ Basic Requirements ]

Input a grammatically correct, variable-free integer expression from the keyboard in the form of a sequence of characters, and use the operator precedence relationship given in Table 3.1 of the textbook to evaluate the four mixed arithmetic expressions, and imitate the textbook example 3 - 1 Demonstrate the change process of operator stack , operand stack , input characters and main operations during evaluation .

[implementation hints ]

( 1 ) Set the operator stack and the operand stack to assist in analyzing the operator precedence relationship;

( 2 ) while reading the sequence of expressions, complete the identification processing of the operand (integer) and the corresponding operation;

( 3 ) While identifying the operand, convert its character sequence form into integer form;

( 4 ) Output the operator stack, operand stack, input characters and the contents of the main operation at the appropriate position in the program

First understand the following arithmetic precedence:

designing process:

Data Type Definition:

1. Sequence stack definition

typedef struct sqstack
{//Sequence stack structure
	elemtype stack[MAXSIZE];
	int top;
}sqstsck;

2. Global variable definition:

char ch[8]={'+' , '-' , '*' , '/' ,'(' , ')' , '#'}; //Convert symbols to character arrays
int f1[7]={3,3,5,5,1,6,0}; //Element priority in stack
int f2[7]={2,2,4,4,6,1,0}; //Priority of elements outside the stack
int n=0;

Source program:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<ctype.h> //Determine whether it is the header file of the character function
#define MAXSIZE  100
typedef int elemtype;
char ch[8]={'+' , '-' , '*' , '/' ,'(' , ')' , '#'}; //Convert symbols to character arrays
int f1[7]={3,3,5,5,1,6,0}; //Element priority in stack
int f2[7]={2,2,4,4,6,1,0}; //Priority of elements outside the stack
int n=0;
typedef struct sqstack
{//Sequence stack structure
	elemtype stack[MAXSIZE];
	int top;
}sqstsck;

//1. Convert the operator to the corresponding number
elemtype cton(char c)
{
	switch(c)
	{
	case '+':  return 0;
	case '-':  return 1;
	case '*':  return 2;
	case '/':  return 3;
	case '(':  return 4;
    case ')':  return 5;
	default:   return 6;
	}
}

//2. Compare character precedence
char Compare(char c1,char c2)
{
	int i1=cton(c1);
	int i2=cton(c2); //turn characters into numbers
	if(f1[i1]>f2[i2]) return '>'; //find the priority through the original setting
	else if(f1[i1]<f2[i2]) return '<';
	else return '=';
}

//3. Four arithmetic operations
int Operate(elemtype a,elemtype t,elemtype b)
{
	int sum
	switch(t)
	{
	case 0: sum=a+b; break;
	case 1: sum=a-b; break;
	case 2: sum=a*b; break;
	default: sum=a/b;
	}
	return sum;
}

//5. Take the top element of the stack
elemtype Gettop(sqstack s)
{
	if(s.top ==0)
	{
		printf("ERROR,underflow\n");
		return 0;
	}
	else return s.stack[s.top];
}

//6. Initialization
void Initstack(sqstack *s)
{
	s->top =0;
}

//7. Pop the stack
void Pop(sqstack *s,elemtype *x)
{
	if(s->top ==0)  printf("ERROR,underflow!\n");
	else
	{
		*x=s->stack [s->top];
		s->top--;
	}
}

//8. Push the stack
void Push(sqstack *s,elemtype x)
{
	if(s->top ==MAXSIZE-1) printf("ERROR,Overflow!\n");
	else
	{
		s->top++;
		s->stack[s->top]=x;
	}
}

//9. If the stack is empty, return TRUE, otherwise return FASLE
bool StackEmpty(sqstack s)
{
	if(s.top ==0) return true;
	else return false;
}
//4. Main expression function
int EvaluateExpression()
{
	char c;
	int i=0,sum=0;
	int k=1,j=1; //Set the switch variable
	elemtype x,t,a,b;
	sqstack OPTR,OPND;
	Initstack(&OPTR);
	Push(&OPTR,cton('#')); //'#' is pushed onto the stack
	Initstack(&OPND);
	c=getchar();
	while(c!='#'||ch[Gettop(OPTR)]!='#')
	{
		if(isdigit(c)) //Determine whether c is a number
		{
			sum=0;
			while(isdigit(c))
			{
				if(!j) //j is used for string conversion judgment, j is 0 conversion
				{
					sum = sum * 10 ( c - ' 0 ' ) ;
				}
				else sum=sum*10+(c-'0'); //The character c is converted into the corresponding number
				c=getchar();
			}
			Push(&OPND,sum); //The current c is not a number, then convert the previous number string to a decimal number and then press
			j=1;
		}
		else if(k)
		{
			switch(Compare(ch[Gettop(OPTR)],c))
			{
			case '<': Push(&OPTR,cton(c)); //Integer the character and push it into the operator stack
				c=getchar();
				break;
			case '=': Pop(&OPTR,&x); //The top element of the operator stack is popped
				c=getchar();
				break;
			case '>': Pop(&OPTR,&t); //The top element of the operator stack is popped
				Pop(&OPND,&b); //The top element of the operand stack is popped from the stack
				Pop(&OPND,&a); //The top element of the operand stack is popped from the stack
				Push(&OPND,Operate(a,t,b));
				break;
			}
		}
	}
	return (Gettop(OPND));
}



void main()
{
	system("color a");
	int result;

	printf("Please enter the expression '#' to end:\n");
	result=EvaluateExpression();
	printf("The result of the expression is: %d\n",result);
	system("pause");  
}