C语言_简单计算器

C语言_简单计算器

文章目录

• • C语言_简单计算器
  • • 1.问题描述
    • 2.算法描述
    • • 有关想法
      • • a.有限状态自动机
        • b.逆波兰表达式
      • 算法实现
      • 1）准备阶段
      • 2）处理字符串
      • 3）利用token序列计算
      • 4）在main函数中调用计算器函数以实现多组样例测试
    • 3.测试数据和测试结果
    • 4.使用说明
    • • • 关于表达式合法性的规定

1.问题描述

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设计简单计算器，能够处理含有‘ ’、‘-’、‘*’、‘/’、‘（’、‘）’、‘^’、‘ ’和非负数整数的混合运算(即加减乘除，括号运算，幂运算)，检测不合法的运算表达式。

输入输出长度不超过80个字符，包括空格，不包含负数。

输入第一行有1个整数t表示测试样例数，以下是t行，每行1个测试样例。

输出精确到小数点后两位。

交互界面友好，有适当输入输出和错误提示。

2.算法描述

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有关想法

a.有限状态自动机

模型：拥有有限数量状态，每个状态可以在一定条件下迁移到零至多个状态（由输入的字符串决定哪个状态的迁移）

使用：用于处理字符串（有限长度），将数字和运算符打包成一个个token（此过程自动按规则进行），便于接下来的运算处理

流程图如下：

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digit

op

除了digit

除了op

空格

digit

op

初始

s0

s1

s2

b.逆波兰表达式

模型：逆波兰表达式又叫后缀表达式，与日常方便人类阅读的中缀表达式相比，更适合计算机阅读。它没有括号，严格遵循从左到右的计算。主要利用运算符的优先级和栈来实现。

表现如下：（图片来源_百度百科）

使用：本实验中的使用与传统后缀表达式运算有所差别——为了方便地实现运算的目的，并没有完全转化为后缀表达式，而是判断、计算、栈操作同时进行。计算过的运算符直接出栈废弃，并在数字栈中舍弃原来的两个数字压入新的数字，即计算结果。（具体实现代码见 3）“利用token序列进行计算”）

p.s.注意在表达式最后加‘#’号作最低优先级符号

算法实现

1）准备阶段

i.设置枚举类型 NUM，OP

ii.规定结构体token，包括类型（NUM，OP）和元素（数字或运算符）

iii.创建功能函数get_num，从字符串中提取数字

代码实现如下：

代码语言：javascript复制

`//从字符串中提取数字并打包
token get_num(const char buf[], int* buf_idx) {
	token ret_token;
	ret_token.type = NUM;
	ret_token.val = 0;
	while (isdigit(buf[*buf_idx])) {
		ret_token.val *= 10;
		ret_token.val  = (double)buf[*buf_idx] - '0';
		(*buf_idx)  ;
	}
	return ret_token;
}`

iv.创建功能函数compare_op，比较两个运算符的优先级

​ 优先级规定：^, * = /, = -, #(由大到小，设‘#’为最低运算符 用于保证最后一次运算)

代码实现如下：

代码语言：javascript复制

//比较两个运算符的优先级
int compare_op(char op1, char op2) {
	static int op_pirority[128];
	op_pirority['#'] = -1;
	op_pirority[' '] = 2;
	op_pirority['-'] = 2;
	op_pirority['*'] = 3;
	op_pirority['/'] = 3;
	op_pirority['^'] = 4;

	int op1_pri = op_pirority[op1];
	int op2_pri = op_pirority[op2];
	if (op1_pri == op2_pri) {
		return 0;
	}
	else if (op1_pri > op2_pri) {
		return 1;
	}
	else {
		return -1;
	}
}

v.创建功能函数int calculate(char, int, int)，进行简单计算（只有一个运算符）

代码实现如下：

代码语言：javascript复制

double calculate(double num1, double num2, char op) {
	if (op == ' ') {
		return num1   num2;
	}
	else if (op == '-') {
		return num1 - num2;
	}
	else if (op == '*') {
		return num1 * num2;
	}
	else if (op == '/') {
        if (num2 == 0){
            printf("PEn");
            //printf("Error: divisor can't be '0'n")
            flag = 1;
            return 0;//随便返回一个整数
        }
		return num1 / num2;
	}
	else if (op == '^') {
		return pow(num1, num2);
	}
	else {
		printf("PEn");
		flag = 1;
		return 0;//同上
	}
}

2）处理字符串

i.输入：以单个字符行书挨个录入数组，遇到换行停止，末尾加‘’

代码实现如下：

代码语言：javascript复制

//1.输入
	while (1) {
		scanf("%c", buf   buf_cnt);
		if (buf[buf_cnt] == 'n') {
			buf[buf_cnt] = '';
			break;
		}
		buf_cnt  ;
	}
	buf_cnt = 0;

ii.将字符串转换成token序列（跳过空格）

代码实现如下：

代码语言：javascript复制

//2.将字符串转换为Token序列
	char ops[] = { ' ', '-','*', '/','^'};
	while (1) {
		//跳过空格
		if (buf[buf_cnt] == ' ') {
			buf_cnt  ;
			continue;
		}

		//遇到则提取数字
		if (isdigit(buf[buf_cnt])) {
			tokens[token_cnt] = get_num(buf, &buf_cnt);
			token_cnt  ;
			continue;
		}

		//判断是否为运算符
		int mark = 0;//利用mark跳出多层循环
		for (i = 0; i < 5; i  ) {
			if (buf[buf_cnt] == ops[i]) {
				tokens[token_cnt].type = OP;
				tokens[token_cnt].op = buf[buf_cnt];
				token_cnt  ;
				buf_cnt  ;
				mark = 1;
			}
		}

		if (mark == 1) {
			continue;
		}

		if (buf[buf_cnt] == '') {
			break;
		}

		printf("PEn");
		flag = 1;
		//printf("Invalid simple at %d: %c", buf_cnt, buf[buf_cnt]);
		
		return;
	}

3）利用token序列计算

当前token为数字时，检查其是否为第偶数个输入（从0开始计数），若是则将其压入数字栈，否则报错 当前token为操作符时，检查其是否为第奇数个输入，不是则报错 若是，若栈内无操作符，则入栈 则将操作符跟当前操作符栈顶元素进行优先级比较 若栈顶优先级大于等于当前运算符优先级， 则出栈两个数字，出栈一个运算符，进行运算，再把数字压入栈，当前操作符压入栈 若栈顶优先级小于当前运算符优先级， 则将当前运算符压入栈 符号栈最后要入一个 # ，标识为最低级的运算符

代码实现如下：

代码语言：javascript复制

tokens[token_cnt].op = '#';
	double num_stack[30];
	int num_cnt = 0;
	char op_stack[30], op_cnt = 0;
	for (i = 0; i <= token_cnt; i  ) {
		if (tokens[i].type == NUM) {
			if ((i) % 2 == 1) {
				printf("PEn");
                //printf("Here lack an operator!n");
				return ;
			}
			num_stack[num_cnt] = tokens[i].val;
			num_cnt  ;
		}
		else {
			if (i % 2 != 1) {
				printf("PEn");
                //printf("Oh, you let two operaters next to each other!")
				return;
			}
			while (1) {
				if (op_cnt == 0) {
					op_stack[op_cnt] = tokens[i].op;
					op_cnt  ;
					break;
				}
				else {
					if (compare_op(tokens[i].op, op_stack[op_cnt - 1]) <= 0) {//若当前优先级小于栈顶运算符
						if (num_cnt < 2) {
							printf("PEn");
                            //printf("Be short of number!n")
							return;
						}
						num_stack[num_cnt - 2] = calculate(num_stack[num_cnt - 2], num_stack[num_cnt - 1], op_stack[op_cnt - 1]);
						if (flag == 1) return;
						num_cnt--;
						op_cnt--;
					}
					else {//若当前优先级大于栈顶运算符
						op_stack[op_cnt] = tokens[i].op;
						op_cnt  ;
						break;
					}
				}
			}
		}
	}
	printf("result: %.2lfn", num_stack[0]);

4）在main函数中调用计算器函数以实现多组样例测试

3.测试数据和测试结果

---

样例序号	输入	输出
1	3 4	7.00
2	2*3	6.00
3	11/3	3.67
4	3 4215-8/2	119.00
5	3 421 5-8/2	PE
6	3 4a215-8/2	PE
7	3^2	9.00
8	2/0	PE
9	7*2-/1	PE
10	*9-1	PE
11	4-8/2^3	3.00
12	9/3 4-	PE
13	11^2/11-1	10.00
14	3*2^0-1 1/4	2.25
15	103-1 188/22-9*11	947.00

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4.使用说明

---

i)输入表达式前，需先输入要计算表达式的个数

ii)输入不合法的表达式将输出“PEn”

iii)计算结果为两位小数的实数，支持负数结果

iv)幂运算使用符号‘^’

v)可以计算加减乘除和幂运算，除数不能为0

关于表达式合法性的规定

输入字符仅可包含数字，空格， ，-，*，/，^ 16个字符

数字中间不可存在空格（会被认为两个操作数中间没有运算符）

不能连续输入两个运算符（括号除外）

0不能做除数

表达式的两端应该是数字

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