请用C语言代码编写程序，执行按中缀形式输入一个四则运算的表达式，利用优选算法把其转换为后缀表达式输出，并求表达式的值。

好的，下面是使用C语言编写的程序，可以执行按中缀形式输入一个四则运算的表达式，利用优选算法把其转换为后缀表达式输出，并求表达式的值。

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

#define MAX_SIZE 100

typedef struct stack {
    char data[MAX_SIZE];
    int top;
} Stack;

void init(Stack *s) {
    s->top = -1;
}

int is_empty(Stack *s) {
    return s->top == -1;
}

int is_full(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

void push(Stack *s, char c) {
    if (is_full(s)) {
        printf("Error: stack is full!\n");
        exit(1);
    }
    s->data[++s->top] = c;
}

char pop(Stack *s) {
    if (is_empty(s)) {
        printf("Error: stack is empty!\n");
        exit(1);
    }
    return s->data[s->top--];
}

char peek(Stack *s) {
    if (is_empty(s)) {
        printf("Error: stack is empty!\n");
        exit(1);
    }
    return s->data[s->top];
}

int priority(char op) {
    switch (op) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        case '(':
            return 0;
        default:
            printf("Error: invalid operator %c\n", op);
            exit(1);
    }
}

void infix_to_postfix(char *infix, char *postfix) {
    Stack s;
    init(&s);
    int i = 0;
    int j = 0;
    while (infix[i]) {
        if (isdigit(infix[i])) {
            postfix[j++] = infix[i++];
            while (isdigit(infix[i])) {
                postfix[j++] = infix[i++];
            }
            postfix[j++] = ' ';
        } else if (infix[i] == '(') {
            push(&s, infix[i++]);
        } else if (infix[i] == ')') {
            while (peek(&s) != '(') {
                postfix[j++] = pop(&s);
                postfix[j++] = ' ';
            }
            pop(&s);
            i++;
        } else if (isspace(infix[i])) {
            i++;
        } else {
            while (!is_empty(&s) && priority(infix[i]) <= priority(peek(&s))) {
                postfix[j++] = pop(&s);
                postfix[j++] = ' ';
            }
            push(&s, infix[i++]);
        }
    }
    while (!is_empty(&s)) {
        postfix[j++] = pop(&s);
        postfix[j++] = ' ';
    }
    postfix[j] = '\0';
}

int evaluate_postfix(char *postfix) {
    Stack s;
    init(&s);
    int i = 0;
    while (postfix[i]) {
        if (isdigit(postfix[i])) {
            int num = 0;
            while (isdigit(postfix[i])) {
                num = num * 10 + postfix[i] - '0';
                i++;
            }
            push(&s, num);
        } else if (postfix[i] == ' ') {
            i++;
        } else {
            int b = pop(&s);
            int a = pop(&s);
            switch (postfix[i]) {
                case '+':
                    push(&s, a + b);
                    break;
                case '-':
                    push(&s, a - b);
                    break;
                case '*':
                    push(&s, a * b);
                    break;
                case '/':
                    push(&s, a / b);
                    break;
                default:
                    printf("Error: invalid operator %c\n", postfix[i]);
                    exit(1);
            }
            i++;
        }
    }
    return pop(&s);
}

int main() {
    char infix[MAX_SIZE];
    char postfix[MAX_SIZE];
    printf("Please enter an infix expression: ");
    fgets(infix, MAX_SIZE, stdin);
    infix_to_postfix(infix, postfix);
    printf("The postfix expression is: %s\n", postfix);
    printf("The result is: %d\n", evaluate_postfix(postfix));
    return 0;
}

输入一个中缀表达式，程序会将其转换为后缀表达式并输出，然后计算后缀表达式的值并输出。注意，程序假设输入的表达式没有语法错误。