帮我用C语言写一个能将算术运算表达式转换为逆波兰式的函数

好的，关于您的问题，我可以回答。以下是一种将中缀表达式转换为逆波兰式的函数：

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

#define SIZE 50

typedef struct stack {
    char items[SIZE];
    int top;
} stack;

void push(stack* s, char c) {
    if (s->top < SIZE - 1) {
        s->items[++(s->top)] = c;
    }
}

char pop(stack* s) {
    if (s->top >= 0) {
        return s->items[(s->top)--];
    }
    return -1;
}

int is_digit(char c) {
    return (c >= '0' && c <= '9');
}

int is_operator(char c) {
    return (c == '+' || c == '-' || c == '*' || c == '/');
}

int precedence(char c) {
    switch (c) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        default:
            return 0;
    }
}

char* infix_to_postfix(char* expression, int n) {
    stack* s = (stack*) malloc(sizeof(stack));
    s->top = -1;
    char* postfix = (char*) malloc(n + 1);
    int i, j = 0;
    for (i = 0; i < n; i++) {
        if (is_digit(expression[i])) {
            postfix[j++] = expression[i];
        } else if (is_operator(expression[i])) {
            while (s->top >= 0 &&