c语言实现数学表达式

问题遇到的现象和发生背景

尝试实现一个c语言的功能，需要输入一个表达式输出对应结果，包括加减乘除乘方，中小括号。

用代码块功能插入代码，请勿粘贴截图
#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#define ARGC_CNT 2
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 10
#define OK 1
#define OVERFLOW 0
#define ERROR 0

typedef  double Status;
typedef  double SElemType;    //操作数
typedef  char status;
typedef char selemType;    //运算符

typedef struct {
    SElemType* base;
    SElemType* top;
    int stacksize;
}SqStack;//操作数栈

typedef struct {
    selemType* base;
    selemType* top;
    int stacksize;
}sqStack;//运算符栈

char prior[10][10] = {
{'>','>','<','<','<','<','>','<','>','>'},
{'>','>','<','<','<','<','>','<','>','>'},
{'>','>','>','>','<','<','>','<','>','>'},
{'>','>','>','>','<','<','>','<','>','>'},
{'>','>','>','>','>','<','>','<','>','>'},
{'<','<','<','<','<','<','=','>','?','?'},
{'>','>','>','>','>','?','>','?','>','>'},
{'<','<','<','<','<','<','?','<','=','?'},
{'>','>','>','>','>','?','<','?','>','>'},
{'<','<','<','<','<','<','?','<','?','='} };

char OPSET[10] = { '+','-','*','/','^','(',')','[',']','#'};

Status InitStack(SqStack* s)//初始化操作数栈 
{
    s->base = (SElemType*)malloc(STACK_INIT_SIZE * sizeof(SElemType));
    if (!s->base)
        exit(OVERFLOW);
    s->top = s->base;
    s->stacksize = STACK_INIT_SIZE;
    return OK;
}

status initStack(sqStack* s)//初始化运算符栈 
{
    s->base = (selemType*)malloc(STACK_INIT_SIZE * sizeof(selemType));
    if (!s->base)
        exit(OVERFLOW);
    s->top = s->base;
    s->stacksize = STACK_INIT_SIZE;
    return OK;
}

Status Push(SqStack* s, SElemType c)//入栈 
{
    if ((s->top - s->base) >= s->stacksize)
    {
        s->base = (SElemType*)realloc(s->base, (s->stacksize + STACKINCREMENT) * sizeof(SElemType));
        if (!s->base)
            exit(OVERFLOW);
        s->stacksize += STACKINCREMENT;
    }
    *(s->top)++ = c;
    return OK;
}

status push(sqStack* s, selemType c)//入栈 
{
    if ((s->top - s->base) >= s->stacksize)
    {
        s->base = (selemType*)realloc(s->base, (s->stacksize + STACKINCREMENT) * sizeof(selemType));
        if (!s->base)
            exit(OVERFLOW);
        s->stacksize += STACKINCREMENT;
    }
    *(s->top)++ = c;
    return OK;
}

Status GetTop(SqStack* s)//取栈顶元素 
{
    SElemType e;
    if (s->base == s->top)
    {
        printf("ERROR_02");//操作数栈为空，即重复运算符
        exit(0);
    }
    e = *(s->top - 1);
    return e;
}

status getTop(sqStack* s)//取栈顶元素 
{
    selemType e;
    if (s->base == s->top)
    {
        printf("ERROR_02");//运算符栈为空
        exit(0);
    }
    e = *(s->top - 1);
    return e;
}

int In(char c)//判断是否为运算符 
{
    if (c=='+'||c=='-'||c=='*'||c=='/'||c=='^'||c=='('||c==')'||c=='['||c==']')
    {
        return OK;
    }
    else
    { 
        return ERROR;
    }
    
}

status precede(char a, char b)//判断优先级函数 
{
    int i = 0, j = 0;
    while (a != OPSET[i]) { i++; }
    while (b != OPSET[j]) { j++; }
    if (prior[i][j] == '?') {
        printf("ERROR_02");//输入表达式错误
        exit(0);
    }
    return prior[i][j];
}

Status Pop(SqStack* s)//脱括号函数 
{
    SElemType e;
    if (s->base == s->top)
        return ERROR;
    e = *--(s->top);
    return e;
}

status pop(sqStack* s) 
{
    int e;
    if (s->base == s->top)
        return ERROR;
    e = *--(s->top);
    return e;
}

Status Opereta(double a, char b, double c)//运算函数 
{
    switch (b)
    {
    case '+':
        return a + c;
    case '-':
        return a - c;
    case '*':
        return a * c;
    case '/':
        if(c==0)
        {
            printf("ERROR_03");//除数为0
            exit(0);
        }
        else 
        {
            return a / c;
        }
    case '^':
        return pow(a, c);
    }
}

int Check(char* S) {
    int i;
    for (i = 0; i < strlen(S); i++) {
        if (S[i] >= '0' && S[i] <= '9')continue;
        if (S[i] == '(' || S[i] == ')' || S[i]=='[' || S[i]==']' || S[i] == '*' || S[i] == '/' || S[i] == '+' || S[i] == '-' || S[i] == '.')continue;
        return ERROR;
    }
    return OK;
}//检查函数,用来检查输入的表达式的操作数和运算符的合法性

double EvaluateExpression(char* MyExpression)
{//算术表达式求值的算符优先算法。
    if (Check(MyExpression) == 0) {
        printf("ERROR_02");//输入的表达式不合法
        exit(0);
    }
 //设OPTR和OPND分别为运算符栈和运算数栈
    sqStack OPTR;//运算符栈，字符元素 
    SqStack OPND;//运算数栈，实数元素 
    char TempData[100];
    char DR[2];
    DR[1] = '\0';
    double data, a, b;
    int l = 0;
    char theta;
    initStack(&OPTR);
    push(&OPTR, '#');
    InitStack(&OPND);
    TempData[0] = '\0';
    while (MyExpression[l]!='#'||getTop(&OPTR)!='#')
    {
        if (!In(MyExpression[l]))//不是运算符则进栈 
        {
            DR[0]=MyExpression[l];
            strcat(TempData,DR);
            l++;
            if (In(MyExpression[l]))//是运算符时 
            {
                data = atof(TempData);
                Push(&OPND, data);
                TempData[0] = '\0';
            }
        }
        else
        {
            switch (precede(getTop(&OPTR),MyExpression[l]))
            {
            case '<':
                push(&OPTR, MyExpression[l]);
                l++;
                break;
            case '=':
                pop(&OPTR);
                l++;
                break;
            case '>':
                a = Pop(&OPND);
                b = Pop(&OPND);
                theta = pop(&OPTR);
                Push(&OPND, Opereta(b, theta, a));
                break;
            }
        }
    }
    return GetTop(&OPND);
}


int main()
{
   char*argv;
   scanf("%s",argv);
    printf("%g", EvaluateExpression(argv));
    return 0;
}
运行结果及报错内容

输入1+2，结果为栈缓冲区溢出

我的解答思路和尝试过的方法

输入后，判断是否有非法字符，如果合法进行运算。依靠栈堆叠来进行，同时判断运算符优先级。最后如果可以正常输出结果则输出，否则输出"ERROR"。

我想要达到的结果

能够正常实现功能，首先希望可以定位出现问题的位置。