后缀表达式求值

最新推荐文章于 2023-03-07 10:38:22 发布

原创最新推荐文章于 2023-03-07 10:38:22 发布 · 569 阅读

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本文介绍了一个使用C语言编写的程序，该程序能够解析并计算后缀表达式的值。通过堆栈结构来处理运算符和操作数，实现了对加、减、乘、除等基本算术运算的支持。

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设计一个程序，输入一个后缀表达式，然后利用堆栈结构的原理，实现求值

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int judgeSymbol(char symbol)
{
	int answer = 317;
	switch (symbol)
	{
	case '+':
		answer = 1; break;
	case '-':
		answer = 2; break;
	case '*':
		answer = 3; break;
	case '/':
		answer = 4; break;
	default:
		break;
	}

	return answer;
}
int main(int argc, char **argv)
{
	int i = 0;
	double mathsAnswer[21] = { '\0' };
	char problem[71] = { '\0' };
	int judge = 666;
	gets_s(problem,70);
	int j = 0;
	int len = strlen(problem);
	for (i = 0; i < len; i++)
	{
		if (problem[i] == ' ')
			continue;//当读取到空格时，抛弃空格
		else if (problem[i] >= '0'&&problem[i] <= '9')
		{
			mathsAnswer[j++] = atof(&problem[i]);
			while (1)
			{
				i++;
				if (problem[i] == ' ')
				{
					i--;
					break;
				}
			}
		}
		else if ((problem[i] == '+' && (problem[i + 1] >= '0'&&problem[i + 1] <= '9')) || (problem[i] == '-'&&problem[i + 1] >= '0'&&problem[i + 1] <= '9'))
		{
			mathsAnswer[j++] = atof(&problem[i]);
			while (1)
			{
				i++;
				if (problem[i] == ' ')
				{
					i--;
					break;
				}
			}
		}
		else
		{
			judge = judgeSymbol(problem[i]);
			if (judge == 1)
			{
				j--;
				mathsAnswer[j - 1] = mathsAnswer[j - 1] + mathsAnswer[j];
				mathsAnswer[j] = '\0';//清空栈顶以上的垃圾值
			}
			else if (judge == 2)
			{
				j--;
				mathsAnswer[j - 1] = mathsAnswer[j - 1] - mathsAnswer[j];
				mathsAnswer[j] = '\0';//清空栈顶以上的垃圾值
			}
			else if (judge == 3)
			{
				j--;
				mathsAnswer[j - 1] = mathsAnswer[j - 1] * mathsAnswer[j];
				mathsAnswer[j] = '\0';//清空栈顶以上的垃圾值
			}
			else if (judge == 4)
			{
				j--;
				if (mathsAnswer[j] == 0)
				{
					printf("ERROR");
					return 0;
				}
				mathsAnswer[j - 1] = mathsAnswer[j - 1] / mathsAnswer[j];
				mathsAnswer[j] = '\0';//清空栈顶以上的垃圾值
			}
			judge = 666;
		}
	}
	if (mathsAnswer[1] != '\0')
		printf("ERROR");
	else
		printf("%.1f\n", mathsAnswer[0]);

	return 0;
}