大一C语言结课设计之《简单计算器》

/*===============================================*\
**	设计目的：简单计算器，计算形如10*(20.2-30.6)+5.0/2的表达式值
**	简要原理：中缀式转逆波兰式（后缀式）
**	IDE：Dev-Cpp 4.9.9.2
**	注意事项：括号必须是英文状态的
**	时间： 2014-6-17
\*===============================================*/

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define maxn 1000
char buf[maxn], str[maxn], signStack[maxn], ch[2];
int len, id, idSign, idAns, i, n;
double ans[maxn];

void checkSign(char sign){
	if(sign == '(') signStack[idSign++] = sign;
	else if(sign == '*' || sign == '/'){
		while(idSign && (signStack[idSign-1] == '*' || signStack[idSign-1] == '/'))
			str[id++] = signStack[--idSign];
		signStack[idSign++] = sign;
	}else if(sign == ')'){
		while(signStack[idSign-1] != '(') str[id++] = signStack[--idSign];
		--idSign;
	}else{
		while(idSign && signStack[idSign-1] != '(') str[id++] = signStack[--idSign];
		signStack[idSign++] = sign;
	}
	str[id++] = ' ';
}

int check(double a, char sign){
	if(a == 0 && sign == '/'){
		printf("除数不能为0!，程序结束。\n");
		system("pause");
		exit(EXIT_FAILURE);
	}
	return 1;
}

double cal(double a, double b, char sign){
	switch(sign){
		case '+': return a + b;
		case '-': return a - b;
		case '*': return a * b;
		case '/': return a / b;
	}
}

int main(){	

	printf("\t\t\t\t简单计算器\n\n");
	printf("\t\t   ======================================\n");
	printf("\t\t  ||  支持+ - * / ( ) 等符号和负数输入  ||\n");
	printf("\t\t   ======================================\n");
	printf("\t\t\t\t\t\t原创作者：邱正钢\n");
	printf("\t\t\t\t\t\t完毕时间：2014年6月17日\n");
	printf("\t\t\t\t\t\tEmail：2276479303@qq.com\n");

	printf("请输入一个表达式，如 -10*(20.2-30.6)+5.0/2\n以回车结束:\n");
	do{
		gets(buf); id = idSign = idAns = 0;
		if((len = strlen(buf)) == 0) continue;

		int flag = 0; //这个是用来推断'-'字符是负号还是减号的。1表示数字，0表示 +*/(
		/*推断原理：假设第一次读取就碰到-，那么一定是负号，若近期一次读取是+*-/(那么
			也一定是负号，假设是右括号或数字那么是减号*/

		for(i = 0; i < len; ++i){
			if(buf[i] == ' ') continue;
			if(buf[i] >= '0' && buf[i] <= '9' || buf[i] == '.' || buf[i] == '-' && !flag){
				 str[id++] = buf[i]; flag = 1;
			}
			else {
				str[id++] = ' '; checkSign(buf[i]);
				if(buf[i] != ')') flag = 0;
				else flag = 1;
			}
		}
		while(idSign) str[id++] = signStack[--idSign];

		for(i = 0, n = 0; i < id; ++i){
			if(str[i] == ' ') continue;
			/*用添加空格的方式推断‘-’字符是负号还是减号，若‘-’后面挨着数字或小数点，那么一定是负号*/
			if(str[i] >= '0' && str[i] <= '9' || str[i] == '.' || str[i] == '-'
				&& (str[i+1] >= '0' && str[i+1] <= '9' || str[i+1] == '.')){
				sscanf(str + i, "%lf%n", &ans[idAns++], &n);
				i += n - 1;  continue;
			}else if(check(ans[idAns-1], str[i])){
				ans[idAns-2] = cal(ans[idAns-2], ans[idAns-1], str[i]);
				--idAns;
			}
		}
		printf("结果是 %.2lf\n", ans[0]);
		printf("输入Y 继续, N 退出: ");
		scanf("%s", ch);
	}while(ch[0] == 'Y' || ch[0] == 'y');
	printf("感谢您的使用！再见\n");
	system("pause");
	return 0;
}