C语言实现的简单的逆波兰计算器

前后调试了好几天，类型转换问题颇为麻烦，无赖使用两种不同类型的栈解决，但程序目前还有两个问题：

1.目前不能进行float类型的计算；

2.不能进行多位数的运算。

实际上就是目前只能进行个位数的带括号的加减乘除运算，目前只能写到这一步，期待后续改进。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define CAPACITY 50
typedef struct stack
{
    int top;
    char arr[CAPACITY];
}*pStack,stack;
pStack initStack1()
{
    pStack st = (pStack)malloc(sizeof(stack));
    st->top = 0;
    return st;
}
int stackEmpty1(pStack st)
{
    return(st->top <= 0);
}
int stackFull1(pStack st)
{
    return(st->top >= CAPACITY);
}
int pushStack1(pStack st, char a)
{
    if(stackFull1(st))
        {
            printf("栈满！\n");
            return 0;
        }
    else
        {
            st->arr[st->top++] = a;
            return 1;
        }
}
char popStack1(pStack st)
{
    char a;
    if(stackEmpty1(st))
    {
        printf("栈空!\n");
        return 0;
    }
    else
    {
        a = st->arr[--st->top];
        return a;
    }
}
char getTop1(pStack st)
{
    if(!stackEmpty1(st))
        return st->arr[st->top-1];
    else
        printf("Empty stack");
    return 0;
}
typedef struct
{
    int top;
    int data[CAPACITY];
}*dataStack;
dataStack initStack2()
{
    dataStack st = (dataStack)malloc(sizeof(dataStack));
    st->top = 0;
    return st;
}
int stackEmpty2(dataStack st)
{
    return(st->top <= 0);
}
int stackFull2(dataStack st)
{
    return(st->top >= CAPACITY);
}
int pushStack2(dataStack st, int data)
{
    if(stackFull2(st))
        {
            printf("栈满！\n");
            return 0;
        }
    else
        {
            st->data[st->top++] = data;
            return 1;
        }
}
int popStack2(dataStack st)
{
    int data;
    if(stackEmpty2(st))
    {
        printf("栈空!\n");
        return 0;
    }
    else
    {
        data = st->data[--st->top];
        return data;
    }
}
int getTop2(dataStack st)
{
    if(!stackEmpty2(st))
        return st->data[st->top-1];
    else
        printf("Empty stack");
    return 0;
}
int priority(char op)
{
    switch(op)
    {
    case '*':
    case '/':return 2;
    case '+':
    case '-':return 1;
    default: return 0;
    }
}
int myAtoi(char a)
{
    return a-'0';
}
int main(void)
{
    int p = 0;
    int i = 0;
    int j;
    int result, tmp1, tmp2;
    char ch, temp;
    char infix[CAPACITY] = "\0";
    char output[CAPACITY];
 
    pStack st = initStack1();
    dataStack st2 = initStack2();
	printf("请输入中缀表达式：");
	scanf("%s", &infix);
    
	printf("后缀表达式为：");
    for(ch = infix[p];ch !='\0';ch = infix[++p])
    {
        switch(ch)
        {
        case '(':pushStack1(st,ch);break;
        case ')':
            {
                while(getTop1(st) != '(')
                    {
                        output[i++] = getTop1(st);
                        printf("%c", popStack1(st));
                    }
                popStack1(st);
                break;
            }
        case '+':
        case '-':
        case '*':
        case '/':
            {
                while(!stackEmpty1(st))
                    {
                        temp = getTop1(st);
                        if(priority(temp) >= priority(ch))
                        {
                            output[i++] = temp;
                            printf("%c", popStack1(st));
                        }
                        else
                        {
                            break;
                        }
                    }
                pushStack1(st, ch);
                break;
            }
        default:output[i++] = ch,printf("%c",ch);
        }
    }
    while(!stackEmpty1(st))
    {
        temp = popStack1(st);
        output[i++] = temp;
        if(temp == '(')
        {
            printf("表达式有误！");
            exit(EXIT_FAILURE);
        }
        printf("%c", temp);
		printf("\n");
    }
    output[i] = '\0';
    free(st);
    for(j=0;j<i;j++)
    {
        switch(output[j])
        {
            case'+':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp1 + tmp2;pushStack2(st2, result);break;
            case'-':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 - tmp1;pushStack2(st2, result);break;
            case'*':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 * tmp1;pushStack2(st2, result);break;
            case'/':tmp1 = popStack2(st2);tmp2 = popStack2(st2);result = tmp2 / tmp1;pushStack2(st2, result);break;
            default:pushStack2(st2,myAtoi(output[j]));break;
        }
    }
    printf("输出结果为: %d\n", getTop2(st2));

    return 0;
}

两种堆栈可以用泛型实现共用函数，目前还不清楚怎么实现。