评估后缀表达式时不需要的符号结果

Question

I'm trying to make a C program to evaluate postfix expressions and when doing so an unwanted symbol is being printed on the screen for the input 45+. PS Please tell me the mistake (except of that gets() I am studying right now how to use fgets())

// to Evaluate a postfix expression
#include<stdio.h>
#include<conio.h>
int is_operator(char);
void answer();
char stack[100];
int top =-1;
void push(char);
char pop();
void main()
{
char postfix[100],item;
int i=0;
clrscr();
printf("Enter Postfix Expression");
gets(postfix);
while(postfix[i]!='\0')
    {
    item=postfix[i];
    if(is_operator(item)==2)
        {
        push(item);
        }
    if(is_operator(item)==1)
        {
        char op;
        int n1,n2,n3;
        op=item;
        n1=pop();
        n2=pop();
        switch(op)
        {
            case '+':
            n3=n1+n2;
            case '-':
            n3=n1-n2;
            case '*':
            n3=n1*n2;
            case '/':
            n3=n1/n2;
        }
        push(n3);
        }
    i++;
    }//end while
answer();
getch();
}
void push(char c)
{
top++;
stack[top]=c;
}
char pop()
{
char c;
c=stack[top];
top--;
return(c);
}
int is_operator(char i)
{
char ch=i;
if(ch=='+'||ch=='-'||ch=='*'||ch=='/')
    {
    return(1);
    }
else
    {
    return(2);
    }
}
void answer()
{
char ans;
ans=stack[top];
printf("Answere is %c",ans);
}

Answer 1

There are a lot of mistakes in your code.Try to properly type cast.

Go through comments to understand the mistakes.

Go through this for understanding character pointer and arrays.

// to Evaluate a postfix expression

#include<stdio.h>



int is_operator(char);
void answer();
int stack[100];//Use integer array since operands are integer
int top =-1;
void push(int);//Arguments changed to integer type since the stack is integer 
int pop(); //Return type to integer
void main()
{
char* postfix;//Use character pointer for iterating through loop smoothly
 int item;
int i=0;

printf("Enter Postfix Expression");
gets(postfix);
char c;

while(*postfix!='\0')
    {
    c=*postfix;
    if(is_operator(c)==2)
        {
        push((c-'0')); //Converting char to int before pushing it into the stack
        }
    if(is_operator(c)==1)
        {


        char op;
        int n1,n2,n3;
        op=*postfix;
        n1=pop();
        n2=pop();
        switch(op)
        {
            case '+':
            n3=n1+n2;
            break;
            case '-':
            n3=n1-n2;
            break;
            case '*':
            n3=n1*n2;
            break;
            case '/':
            n3=n1/n2;
            break;
        }
        push(n3);
        }
    postfix++;
    }//end while
answer();

}
void push(int c)
{
top++;
stack[top]=c;
}
int pop()
{
int c;
c=stack[top];
top--;
return(c);
}
int is_operator(char i)
{
char ch=i;
if(ch=='+'||ch=='-'||ch=='*'||ch=='/')
    {
    return(1);
    }
else
    {
    return(2);
    }
}
void answer()
{
char ans;
ans=stack[top];
printf("Answere is %d",ans);
}  

I hope it is helpful....

Answer 2

The problems I see with your code are: your switch() is missing break statements on the individual case clauses (and a default case might be nice too); when you push your non-operators (aka single digit numbers) on the stack, you push them as character codes rather then converting them to numbers so the math doesn't make sense; you're not properly handling the order of non-communitive operations like subtraction and division (use Unix dc command as a comparison tool); finally, don't reinvent booleans. Below is a rework of your code with the above changes and some style adjustments:

// Evaluate a postfix expression

#include <ctype.h>
#include <stdio.h>
#include <stdbool.h>

char stack[100];
int top = -1;

void push(char);
char pop(void);
bool is_operator(char);
void answer(void);

void push(char c)
{
    stack[++top] = c;
}

char pop()
{
    return stack[top--];
}

bool is_operator(char op)
{
    return (op == '+' || op == '-' || op == '*' || op == '/');
}

void answer()
{
    printf("Answer is %d\n", stack[top]);
}

int main()
{
    char item, postfix[100];
    int i = 0;

    printf("Enter Postfix Expression: ");

    gets(postfix);

    while ((item = postfix[i++]) != '\0')
    {
        if (is_operator(item))
        {
            char n1 = pop();
            char n2 = pop();
            char n3 = 0;

            switch (item)
            {
                case '+':
                    n3 = n1 + n2;
                    break;
                case '-':
                    n3 = n2 - n1;
                    break;
                case '*':
                    n3 = n1 * n2;
                    break;
                case '/':
                    n3 = n2 / n1;
                    break;
            }

            push(n3);
        } else if (isdigit(item)) {
            push(item - '0');
        }
    } // end while
    answer();

    return 0;
}

Example (note this evaluator only operates on single digit numbers):

> ./a.out
Enter Postfix Expression: 6 4 - 7 * 1 +
Answer is 15
> dc
6 4 - 7 * 1 + p
15