C语言_简单计算器

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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

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

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）

img

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("PE\n");

//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("PE\n");

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‘\0’

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] = '\0';

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] == '\0') {
		break;
	}

	printf("PE\n");
	flag = 1;
	//printf("Invalid simple at %d: %c", buf_cnt, buf[buf_cnt]);
	
	return;
}</code></pre></div></div><h5 id="bhqq6" name="3%EF%BC%89%E5%88%A9%E7%94%A8token%E5%BA%8F%E5%88%97%E8%AE%A1%E7%AE%97">3）利用token序列计算</h5><p>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("PE\n");

//printf("Here lack an operator!\n");

return ;

}

num_stack[num_cnt] = tokens[i].val;

num_cnt++;

}

else {

if (i % 2 != 1) {

printf("PE\n");

//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("PE\n");

//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: %.2lf\n", 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
23
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
72-/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
32^0-1+1/4
2.25
15
103-1+188/22-911
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“PE\n”
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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