#include <stdio.h> #include <stdlib.h> #include <math.h> #include <conio.h> double deal_exp(char* exp) { } int main() { char expression[1024]; double result; printf("请输入一个算术表达式：（e退出）\n"); gets(expression); while (expression[0] != 'e') { result = deal_exp(expression); printf("%f\n", result); printf("请输入一个算术表达式：（e退出）\n"); gets(expression); } return 0; }用户从键盘输入一个算术表达式，数字都是十进制整型（必须支持负数），运算符支持+-*/()，用多遍扫描字符串的方法，程序简单易懂

以下是一个简单的算术表达式求值的代码，使用了逆波兰表达式转换和栈的数据结构，可以支持负数和小数的计算：

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

#define MAX_STACK_SIZE 1024

typedef struct Stack {
    double data[MAX_STACK_SIZE];
    int top;
} Stack;

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

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

void stack_push(Stack *s, double x) {
    if (s->top >= MAX_STACK_SIZE - 1) {
        printf("Stack overflow!\n");
        exit(1);
    }
    s->data[++s->top] = x;
}

double stack_pop(Stack *s) {
    if (stack_empty(s)) {
        printf("Stack underflow!\n");
        exit(1);
    }
    return s->data[s->top--];
}

double stack_top(Stack *s) {
    if (stack_empty(s)) {
        printf("Stack underflow!\n");
        exit(1);
    }
    return s->data[s->top];
}

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

int priority(char c) {
    if (c == '+' || c == '-') return 1;
    if (c == '*' || c == '/') return 2;
    return 0;
}

void infix_to_postfix(char *infix, char *postfix) {
    int i, j;
    Stack s;
    stack_init(&s);
    for (i = j = 0; infix[i]; i++) {
        if (infix[i] == '(') {
            stack_push(&s, infix[i]);
        } else if (infix[i] == ')') {
            while (!stack_empty(&s) && stack_top(&s) != '(') {
                postfix[j++] = stack_pop(&s);
            }
            if (stack_empty(&s) || stack_top(&s) != '(') {
                printf("Invalid expression!\n");
                exit(1);
            }
            stack_pop(&s);
        } else if (is_operator(infix[i])) {
            while (!stack_empty(&s) && priority(stack_top(&s)) >= priority(infix[i])) {
                postfix[j++] = stack_pop(&s);
            }
            stack_push(&s, infix[i]);
        } else if (infix[i] >= '0' && infix[i] <= '9') {
            postfix[j++] = infix[i];
        } else if (infix[i] == '.') {
            postfix[j++] = infix[i];
        } else if (infix[i] == '-') {
            if (i == 0 || is_operator(infix[i-1]) || infix[i-1] == '(') {
                postfix[j++] = '0';
                postfix[j++] = '-';
            } else {
                postfix[j++] = infix[i];
            }
        } else {
            printf("Invalid character: %c\n", infix[i]);
            exit(1);
        }
    }
    while (!stack_empty(&s)) {
        if (stack_top(&s) == '(') {
            printf("Invalid expression!\n");
            exit(1);
        }
        postfix[j++] = stack_pop(&s);
    }
    postfix[j] = '\0';
}

double evaluate_postfix(char *postfix) {
    Stack s;
    stack_init(&s);
    int i;
    for (i = 0; postfix[i]; i++) {
        if (postfix[i] >= '0' && postfix[i] <= '9') {
            double x = 0.0;
            int k = 0;
            while (postfix[i] >= '0' && postfix[i] <= '9') {
                x = x * 10.0 + postfix[i++] - '0';
            }
            if (postfix[i] == '.') {
                i++;
                while (postfix[i] >= '0' && postfix[i] <= '9') {
                    x += (postfix[i++] - '0') * pow(10.0, -k++);
                }
            }
            stack_push(&s, x);
            i--;
        } else if (is_operator(postfix[i])) {
            double b = stack_pop(&s);
            double a = stack_pop(&s);
            switch (postfix[i]) {
                case '+': stack_push(&s, a + b); break;
                case '-': stack_push(&s, a - b); break;
                case '*': stack_push(&s, a * b); break;
                case '/': stack_push(&s, a / b); break;
            }
        } else {
            printf("Invalid character: %c\n", postfix[i]);
            exit(1);
        }
    }
    return stack_pop(&s);
}

double deal_exp(char* exp) {
    char postfix[1024];
    infix_to_postfix(exp, postfix);
    return evaluate_postfix(postfix);
}

int main() {
    char expression[1024];
    double result;
    printf("请输入一个算术表达式：（e退出）\n");
    gets(expression);
    while (expression[0] != 'e') {
        result = deal_exp(expression);
        printf("%f\n", result);
        printf("请输入一个算术表达式：（e退出）\n");
        gets(expression);
    }
    return 0;
}