在c ++中评估后缀表达式

问题描述

由于某种原因，该程序运行但编译器正在终止并将我发送到调试和终止。



任何想法为什么？我被卡住了。



我在代码中评论了它应该做什么。我可以废弃这一切，只是尝试别的东西，并相信我，我即将哈哈哈哈，但我真的想知道为什么编译器正在把它取回来

For some reason, this program runs but compiler is terminating and sends me to debug and termination.

Any idea why? I'm stuck like chuck.

I commented in the code what it's supposed to do. I can scrap this whole thing and just try something else, and trust me, I'm about to ha ha ha, but I really want to know why the compiler is puking it back out

#include<iostream>
#include<conio.h>
#include<ctype.h>

using namespace std;

/*
	The program will evaluate a postfix expression that contains digits and operators.
	The program tries to simulate the microprocessor execution stack or evaluation
	of expression.
 */

//The class performing the evaluation
class Evaluation {
	public:
		int st[50];
		int top;
		char str[50];
		Evaluation() {
			top = -1;
		}

		//function to push the item
		void push(int item) {
			top++;
			st[top] = item;
		}

		//function to pop an item
		int pop() {
			int item = st[top];
			top--;
			return item;
		}

		//function to perform the operation depending on the operator.
		int operation(int a,int b,char opr) {
			switch(opr) {
				case '+':return a+b;
				case '-':return a-b;
				case '*':return a*b;
				case '/':return a/b;
				default: return 0;
			}
		}

		int calculatePostfix();
};

//This is the function that calculates the result of postfix expression.
int Evaluation::calculatePostfix() {
	int index = 0;
	while(str[index]!='#') {
		if(isdigit(str[index])) {
			push(str[index]-'0');
		}
		else {
			int x = pop();
			int y = pop();
			int result = operation(x,y,str[index]);
			push(result);
		}
		index++;
	}
	return pop();
}

/*
	main function that reads the postfix expression and	that prints the result.

	The input expression should be ending with a number
	An example input expression would be:
	123*+
	Its output will be 7.

*/
int main() {
	void clrscr();
	Evaluation eval;
	cout << "Enter the postfix: ";
	cin >> eval.str;
	int result = eval.calculatePostfix();
	cout << "the result is " << result;
	getch();
}

推荐答案

while(str[index]!='#') {

表示它会循环遍历 str ，直到找到＃，如果字符串中没有，那么它将失败。



尝试输入此内容;

means it will loop over str until it finds a #, if that is not present in the string then it will fail.

Try inputting this;

123*+#

希望这会有所帮助，

Fredrik

Hope this helps,
Fredrik

这篇关于在c ++中评估后缀表达式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！