使用visual studio2019写C语言代码，用中缀法完成四则运算表达式求解（要求手动输入原始表达式后输出计算结果）。

好的，以下是代码实现：

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

#define MAXSIZE 100 // 定义栈的最大长度

// 定义栈结构体
typedef struct
{
    int data[MAXSIZE];
    int top;
} Stack;

// 初始化栈
void InitStack(Stack *s)
{
    s->top = -1;
}

// 判断栈是否为空
int IsEmpty(Stack *s)
{
    return s->top == -1;
}

// 判断栈是否已满
int IsFull(Stack *s)
{
    return s->top == MAXSIZE - 1;
}

// 入栈
void Push(Stack *s, int x)
{
    if (IsFull(s))
    {
        printf("Stack is full.\n");
        exit(0);
    }
    s->data[++s->top] = x;
}

// 出栈
int Pop(Stack *s)
{
    if (IsEmpty(s))
    {
        printf("Stack is empty.\n");
        exit(0);
    }
    return s->data[s->top--];
}

// 获取栈顶元素
int GetTop(Stack *s)
{
    if (IsEmpty(s))
    {
        printf("Stack is empty.\n");
        exit(0);
    }
    return s->data[s->top];
}

// 判断是否为运算符
int IsOperator(char ch)
{
    return ch == '+' || ch == '-' || ch == '*' || ch == '/';
}

// 获取运算符的优先级
int GetPriority(char ch)
{
    switch (ch)
    {
    case '+':
    case '-':
        return 1;
    case '*':
    case '/':
        return 2;
    default:
        return 0;
    }
}

// 中缀表达式转换为后缀表达式
void InfixToPostfix(char *infix, char *postfix)
{
    Stack s;
    InitStack(&s);

    int i = 0;
    int j = 0;

    while (infix[i] != '\0')
    {
        if (infix[i] >= '0' && infix[i] <= '9')
        {
            postfix[j++] = infix[i++];
            while (infix[i] >= '0' && infix[i] <= '9')
            {
                postfix[j++] = infix[i++];
            }
            postfix[j++] = ' ';
        }
        else if (IsOperator(infix[i]))
        {
            while (!IsEmpty(&s) && GetPriority(GetTop(&s)) >= GetPriority(infix[i]))
            {
                postfix[j++] = Pop(&s);
                postfix[j++] = ' ';
            }
            Push(&s, infix[i++]);
        }
        else if (infix[i] == '(')
        {
            Push(&s, infix[i++]);
        }
        else if (infix[i] == ')')
        {
            while (GetTop(&s) != '(')
            {
                postfix[j++] = Pop(&s);
                postfix[j++] = ' ';
            }
            Pop(&s);
            i++;
        }
        else
        {
            i++;
        }
    }

    while (!IsEmpty(&s))
    {
        postfix[j++] = Pop(&s);
        postfix[j++] = ' ';
    }

    postfix[j] = '\0';
}

// 计算后缀表达式的值
int EvaluatePostfix(char *postfix)
{
    Stack s;
    InitStack(&s);

    int i = 0;

    while (postfix[i] != '\0')
    {
        if (postfix[i] >= '0' && postfix[i] <= '9')
        {
            int num = 0;
            while (postfix[i] >= '0' && postfix[i] <= '9')
            {
                num = num * 10 + postfix[i] - '0';
                i++;
            }
            Push(&s, num);
        }
        else if (postfix[i] == '+')
        {
            int b = Pop(&s);
            int a = Pop(&s);
            Push(&s, a + b);
            i++;
        }
        else if (postfix[i] == '-')
        {
            int b = Pop(&s);
            int a = Pop(&s);
            Push(&s, a - b);
            i++;
        }
        else if (postfix[i] == '*')
        {
            int b = Pop(&s);
            int a = Pop(&s);
            Push(&s, a * b);
            i++;
        }
        else if (postfix[i] == '/')
        {
            int b = Pop(&s);
            int a = Pop(&s);
            Push(&s, a / b);
            i++;
        }
        else
        {
            i++;
        }
    }

    return Pop(&s);
}

int main()
{
    char infix[MAXSIZE];
    char postfix[MAXSIZE];

    printf("请输入中缀表达式：");
    gets(infix);

    InfixToPostfix(infix, postfix);

    printf("后缀表达式为：%s\n", postfix);

    int result = EvaluatePostfix(postfix);

    printf("计算结果为：%d\n", result);

    return 0;
}

输入中缀表达式后，程序会自动将其转换为后缀表达式，并输出计算结果。