C语言利用栈求中缀表达式

Question

#include<stdio.h>
#include<malloc.h> 
#include<stdlib.h>

#define STACK_INIT_SIZE 20
#define STACKINCREMENT 20

typedef struct {
    int* base;
    int* top;
    int stacksize;
}SqStack;

int InitStack(SqStack &S);
int Push(SqStack& S, char e);
int Pop(SqStack& S, char& e);
char GetTop(SqStack S);
int In(char ch); //运算符判断
char Precede(char ch1, char ch2); //优先级判断
int Operate(double a, char theta, double b); //运算方式
int Tz(char* str);// 弹栈 

int main() {
    char str[STACK_INIT_SIZE] = {};//输入的字符串算术表达式
    char ch = 0;
    int output = 0;
    while (true)
    {
        scanf("",&str);//输入
        ch = str[0];
        if (ch == '=')
            break;
        else
        {
            output = Tz(str);
        }
    }
    return 0;
}

int InitStack(SqStack &S) {
    S.base = (int*)malloc(STACK_INIT_SIZE * sizeof(int));
    if (!S.base)
        return 0;
    S.top = S.base;
    S.stacksize = STACK_INIT_SIZE;
    return 1;
}
int Push(SqStack& S, char e) {
    if (S.top - S.base >= S.stacksize) {
        S.base = (int*)realloc(S.base, (S.stacksize + STACKINCREMENT));
        if (!S.base)
            return 0;
        S.top = S.base + S.stacksize;
        S.stacksize += STACKINCREMENT;
    }
    *S.top = e;
    S.top++;
    return 1;
}
int Pop(SqStack& S, char& e){
    if (S.top == S.base)
        return -1;
    --S.top;
    e =* S.top;
    return 1;
}
char GetTop(SqStack S) {
    if (S.base != S.top)
        return *S.top;
    else
        return 0;
}
int In(char ch) {
    if (ch == '+' || ch == '-' || ch == '*'
        || ch == '/' || ch == '(' || ch == ')' || ch == '=')
        return true;
    else
        return false;

}
char Precede(char ch1, char ch2) {
    char cmpResult = 0;//返回的比较结果> < =
    switch (ch1)
    {
    case '?':
        cmpResult = '<';
        break;
    case '+':
        switch (ch2)
        {
        case '+':
            cmpResult = '>';
            break;
        case '-':
            cmpResult = '>';
            break;
        case '*':
            cmpResult = '<';
            break;
        case '/':
            cmpResult = '<';
            break;
        case '(':
            cmpResult = '<';
            break;
        case ')':
            cmpResult = '>';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case '-':
        switch (ch2)
        {
        case '+':
            cmpResult = '>';
            break;
        case '-':
            cmpResult = '>';
            break;
        case '*':
            cmpResult = '<';
            break;
        case '/':
            cmpResult = '<';
            break;
        case '(':
            cmpResult = '<';
            break;
        case ')':
            cmpResult = '>';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case '*':
        switch (ch2)
        {
        case '+':
            cmpResult = '>';
            break;
        case '-':
            cmpResult = '>';
            break;
        case '*':
            cmpResult = '>';
            break;
        case '/':
            cmpResult = '>';
            break;
        case '(':
            cmpResult = '<';
            break;
        case ')':
            cmpResult = '>';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case '/':
        switch (ch2)
        {
        case '+':
            cmpResult = '>';
            break;
        case '-':
            cmpResult = '>';
            break;
        case '*':
            cmpResult = '>';
            break;
        case '/':
            cmpResult = '>';
            break;
        case '(':
            cmpResult = '<';
            break;
        case ')':
            cmpResult = '>';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case '(':
        switch (ch2)
        {
        case '+':
            cmpResult = '<';
            break;
        case '-':
            cmpResult = '<';
            break;
        case '*':
            cmpResult = '<';
            break;
        case '/':
            cmpResult = '<';
            break;
        case '(':
            cmpResult = '<';
            break;
        case ')':
            cmpResult = '=';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case ')':
        switch (ch2)
        {
        case '+':
            cmpResult = '>';
            break;
        case '-':
            cmpResult = '>';
            break;
        case '*':
            cmpResult = '>';
            break;
        case '/':
            cmpResult = '>';
            break;
        case '(':
            cmpResult = ' ';
            break;
        case ')':
            cmpResult = '>';
            break;
        case '=':
            cmpResult = '>';
            break;
        default:
            break;
        }
        break;
    case '=':
        break;
    default:
        break;
    }
    return cmpResult;
}
int Operate(double a, char theta, double b)
{
    int result = 0;
    switch (theta)
    {
    case '+':
        result = a + b;
        break;
    case '-':
        result = a - b;
        break;
    case '*':
        result = a * b;
        break;
    case '/':
        result = a / b;
        break;
    default:
        break;
    }
    return result;
    return result;

}
int Tz(char* str) {
    SqStack s1, s2; //s1=数字 s2=操作符
    int InitStack(SqStack & s1);//初始化s1 s2
    int InitStack(SqStack & s2);
    char ch;
    ch = str[0];
    int i = 0;//扫描字符串的下标
    char theta = 0;
    char a = 0;
    char b = 0;
    char x = 0;
    int i1 = 0;//数字和小数点构成的字符串下标
    char z[10] = {};//存放数字的字符串
    while (true) {
        ch = str[i];//第i个字符
        if (ch == '\0')
            break;
        if (In(ch))//是运算符
        {
            ch = str[i];
            switch (Precede(GetTop(s2), ch))
            {//比较s2的栈顶元素和ch的优先级
            case '<':
                Push(s2, ch); i++; ch = str[i];
                break;
            case '>':
                Pop(s2, theta);//弹出栈顶的运算符
                Pop(s1, b);
                Pop(s1, a);//弹出栈顶的两个运算数
                Push(s1, Operate(a, theta, b));//将运算结果压入OPND栈
                break;
            case '='://OPTR的栈顶元素是"("且ch")"
                Pop(s1, x);
                i++; ch = str[i];
                break;
            default:
                break;
            }
        }
        else if (ch >= '0' && ch <= '9')//数字
        {
            i1 = 0;
            do
            {
                z[i1] = ch;
                i1++;
                i++;
                ch = str[i];
            } while ((ch >= '0' && ch <= '9') || ch == '.');
            z[i1] = 0;
            double d = atof(z);
            Push(s1, d);
        }
        else
        {
            printf("输入错误！\n");
            break;
        }
    }
    return GetTop(s1);
}

错误如下，明明在上面调用了InitStack函数，为啥还提示这个错误呢？s1却不提示

其他地方还有错误的话，恳求改正！

Answer 1

initstack函数调用不是这么写的
写为InitStack(&s1);这种形式试一下