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c++ - Infix to Postfix to Output (Postfix Calculator) 使用堆栈

转载 作者:太空宇宙 更新时间:2023-11-04 14:17:52 24 4
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大家好!我是 C++ 的新手(这里也是 stackoverflow 的新手),我需要各位专家的帮助。

我这里有一个代码,它应该向用户询问中缀表达式,然后将其转换为后缀并输出结果(后缀计算器)。但是,我无法立即将后缀转换为输出,因此一旦它显示后缀表达式,它就会在输出真正的答案之前再次请求后缀表达式(例如 1 2 + 之后有空格)。

没有错误或警告,但是当我运行程序时,计算机在显示后缀表达式后说“file.exe 已停止工作”。所以程序能够正确地将中缀表达式转换为后缀表达式,但在显示输出时仍然存在一些错误。

#include <iostream>
#include <cstring>
#include <cstdlib>

using namespace std;

struct node {
char data;
node *next;
};

node *top=NULL;
node *bottom=NULL;
node *key;
node *last;
node *before_last;


void push (const char Symbol) {
key = new node;
key->data = Symbol;
key->next = top;
top = key;
}

void push_for_output (node* &stack, int key) {
node* newNode = new node;
newNode->data = key;
newNode->next = stack;
stack = newNode;
}

const char pop() {
if (!top) {
cout << "Stack underflow\n" << endl;
return ' ';
}
node* key = top;
top = top->next;
char ch = key->data;
delete key;
return ch;
}

int pop_for_output (node* &stack) {
int key = stack->data;
node* nodeToDelete = stack;
stack = stack->next;
delete nodeToDelete;
return key;
}

bool isOperator (char *token) {
if (strcmp(token, "+") == 0) {
return true;
}
else if (strcmp(token, "-") == 0) {
return true;
}
else if (strcmp(token, "*") == 0) {
return true;
}
else if (strcmp(token, "/") == 0) {
return true;
}
else {
return false;
}
}



const bool is_empty() {
return !top;
}

int postfix(const char *infix) {
char infix_ch[100]={NULL};
char postfix_ch[100]={NULL};
node* stack = NULL;

strcpy(infix_ch,"(");
strcat(infix_ch, infix);
strcat(infix_ch,")");

char symbol[5]={NULL};
char temp[5]={NULL};

for(int i=0; i<strlen(infix_ch); i++) {
symbol[0]=infix_ch[i];

if(symbol[0]=='(')
push(symbol[0]);

else if(symbol[0]==')') {
symbol[0]=pop( );
while(symbol[0]!='(') {
strcat(postfix_ch, symbol);
symbol[0]=pop( );
}
}

else if(symbol[0]=='^' || symbol[0]=='*' || symbol[0]=='/' || symbol[0]=='+' || symbol[0]=='-') {
if(symbol[0]=='*' || symbol[0]=='/') {
temp[0]=pop( );
while(temp[0]=='^' || temp[0]=='*' || temp[0]=='/') {
strcat(postfix_ch, temp);
temp[0]=pop( );
}
push(temp[0]);
}

else if(symbol[0]=='+' || symbol[0]=='-') {
temp[0]=pop( );
while(temp[0]!='(') {
strcat(postfix_ch, temp);
temp[0]=pop( );
}
push(temp[0]);
}
push(symbol[0]);
}

else
strcat(postfix_ch, symbol);
}

cout << "Postfix: " << postfix_ch;

char postfix[80];
cout << "\nEnter postfix expression (include spaces between each operand and/or operator): ";
cin.getline(postfix, 80);
char *tokens = strtok(postfix, " ");
while (tokens != NULL) {
if (isOperator (tokens)) {
int operand2 = pop_for_output(stack);
int operand1 = pop_for_output(stack);
int result;

if (strcmp(tokens, "+") == 0) {
result = operand1 + operand2;
}
else if (strcmp(tokens, "-") == 0) {
result = operand1 - operand2;
}
else if (strcmp(tokens, "*") == 0) {
result = operand1 * operand2;
}
else if (strcmp(tokens, "/") == 0) {
result = operand1 / operand2;
}

push_for_output (stack, result);
}
else {
push_for_output (stack, atoi (tokens));
}
tokens = strtok(NULL, " ");
}
cout << pop_for_output(stack);
system("pause");
return 0;
}


int main( ) {
char infix_values[100]={NULL};
cout << "Enter the infix equation: ";
cin >> infix_values;
postfix(infix_values);
}

我是新手,我真的需要各位专家的帮助。如果你帮我改正我的程序,我将不胜感激。非常感谢,祝您有愉快的一天!

最佳答案

一个可能的问题是 pop_for_output() 函数永远不会像您在 pop() 中那样检查空/NULL 堆栈。如果输入了无效的后缀表达式,或者如果您的解析不正确,您很容易陷入引用 NULL 指针的情况,这可以很好地解释崩溃。

关于c++ - Infix to Postfix to Output (Postfix Calculator) 使用堆栈,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9803859/

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