Item 5--suffix expression

Questions and codes:

/*2015, College of Computer and Control engineering, Yantai University
 * Author: Ersanli
 * Completion Date: October 7, 2015
 * Problem Description: The algorithm of converting an infix expression to the corresponding suffix expression is realized by using the basic operation of the stack in the sqstack.h. For example, input (56-20)/(4+2), output suffix expression:: 56#20#-4#2#+/requires the number to be appended with #.
 * Input Description: infix expression
 * result output: suffix expression */

(1) header file:

#ifndef sqstack_h_included #define sqstack_h_included #define MAXSIZE #define MAXOP 7 typedef int ELEMTYPE;
    typedef struct {elemtype data[maxsize];                int top;                  Stack pointer} sqstack;   The sequence stack type defines the struct//set operator precedence {char ch;   operator int pri; Priority} lpri[]= {' = ', 0},{' (', 1},{' * ', 5},{'/', 5},{' + ', 3},{'-', 3},{') ', 6}}, rpri[]= {' = ', 0},{' (', 6},{' * ', 4},{'/', 4}


, {' + ', 2},{'-', 2},{') ', 1}};    void Initstack (Sqstack *&s);  Initialize stack void Destroystack (Sqstack *&s);     Destroy stack bool Stackempty (Sqstack *s);  Whether the stack is empty int stacklength (Sqstack *s); Returns the number of elements in the stack-the stack length bool Push (Sqstack *&s,elemtype e); into the stack bool POPs (Sqstack *&s,elemtype &e); Out of stack bool GetTop (sqstack *s,elemtype &e);  Take stack top data element void Dispstack (Sqstack *s);
Output stack void multibaseoutput (int number,int base);
int Leftpri (char op);
int Rightpri (char op);
BOOL Inop (char ch);
int precede (char op1,char OP2);

void Trans (char *exp,char postexp[]); #endif//sqstack_h_included

(2) source program:

#include <stdio.h> #include <malloc.h> #include "SqStack.h" void Initstack (Sqstack *&s) {s= (sqstack
    *) malloc (sizeof (sqstack));
s->top=-1; } void Destroystack (Sqstack *&s) {free (s);} int stacklength (Sqstack *s)//Returns the number of elements in the stack-stack length {return (s->top+
1); } bool Stackempty (Sqstack *s) {return (S-&GT;TOP==-1);} bool Push (Sqstack *&s,elemtype e) {if (S->top==ma
    XSIZE-1)//full case, ie overflow on stack return false;
    s->top++;
    s->data[s->top]=e;
return true;
    The case of bool Pop (Sqstack *&s,elemtype &e) {if (s->top==-1)//stack is empty, that is, the stack overflows with return false;
    e=s->data[s->top];
    s->top--;
return true;
    } bool GetTop (Sqstack *s,elemtype &e) {if (s->top==-1)//stack is empty, that is, the stack overflows with return false;
    e=s->data[s->top];
return true;
    } void Dispstack (Sqstack *s)//output stack {int i;
    for (i=s->top;i>=0;i--) printf ("%c", S->data[i]);
printf ("\ n"); } INT Leftpri (char op)//left operator op priority {int i;
for (i=0; i<maxop; i++) if (LPRI[I].CH==OP) return lpri[i].pri;
    } int Rightpri (char op)//Right operator op priority {int i;
for (i=0; i<maxop; i++) if (RPRI[I].CH==OP) return rpri[i].pri; } bool Inop (char ch)//Determine if CH is operator {if (ch== ' (' | | ch== ') ' | | ch== ' + ' | | ch== '-' | | ch== ' * ' | |
    '/') return true;
else return false;
    } int Precede (char Op1,char op2)//op1 and OP2 operator Precedence comparison result {if (Leftpri (OP1) ==rightpri (OP2)) return 0;
    else if (Leftpri (OP1) <rightpri (OP2)) return-1;
else return 1;               } void Trans (char *exp,char postexp[])//Converts an arithmetic expression exp to a suffix expression postexp {sqstack *opstack;                Defines the operator stack int i=0;
    I as subscript elemtype ch of the postexp;   Initstack (Opstack);
    Use initialization stack operation to allocate space for stack, be sure to do Push (opstack, ' = '); while (*exp!= ')//exp expression does not finish scanning when the loop {if (! Inop (*EXP))//numeric characters{while (*exp>= ' 0 ' && *exp<= ' 9 ')//is determined as the number {postexp[i++]
                =*exp;
            exp++;   } postexp[i++]= ' # ';   Use # to identify a value string to end} else//For the case of operator {GetTop (opstack, ch);
                operator switch with top of stack (precede (ch, *exp)) {case-1://top of stack operator low priority: in-stack
                Push (Opstack, *exp);     exp++;
            Continue to scan for other characters break;      Case 0://Only brackets satisfy this condition Pop (opstack, ch);     Will (Stack exp++;
            Continue to scan for other characters break;
                Case 1://rewind stack and output to postexp postexp[i++]=ch;
                Pop (Opstack, ch);
            Break
    }}}//while (*exp!= ') Pop (Opstack, ch);
        while (ch!= ' = ')//At this time exp scan is complete, back to ' = ' until {postexp[i++]=ch;
    Pop (Opstack, ch);    } postexp[i]= ' + '; To givePostexp expression add end-of-identity destroystack (opstack); }

(3) Debug function:

#include <stdio.h>
#include "SqStack.h"
int main ()
{
    char exp[]= "(56-20)/(4+2)"; You can change exp to keyboard input
    char postexp[200];
    Trans (exp,postexp);
    printf ("Infix expression:%s\n", exp);
    printf ("Suffix expression:%s\n", postexp);
    return 0;
}

Operation Result:

Summary of Knowledge points:

Stack......

Learning experience:

The process is still a bit complicated, and there are many places that are not very clear.